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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1451. |
Acidic solution is defined as a solution whose `[H^(o+)] gt [overset(Theta)OH]`. Base solution has `[overset(Theta)OH] gt [H^(o+)]`. During acid-base titrations, `pH` of the mixture will change depending on the amount base added. This variation is shown in the form of graph by making plot as titration curves `100mL` of `1.0 M H_(3)A (K_(a_(1)) = 10^(-3), K_(a_(2)) = 10^(-5), K_(a_(3)) = 10^(-7))` is titrated against `0.1M NaOh`. The titration curve is as follows. What will be the change in `pH` from point `B` to point `C?`A. `2.8`B. `3.2`C. `4.6`D. `0.94` |
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Answer» Correct Answer - D At point `B,Na_(2)HA` is formed, hydrolysis of `HA^(2-)` is considered and hydrolysis of `NaH_(2)A `is supressed `(K_(a_(2))` is considered) due to common ion `[overset(Theta)OH]` effect. `HA^(2-) + H_(2)O hArr H_(2)A^(Theta) + overset(Theta)OH` `H_(2)A^(Theta) + H_(2)O hArr H_(3)A + overset(Theta)OH` At point `B`, salt `Na_(2)HA` (salt of `W_(A)//S_(B))` is formed. So the volume of `NaOH` used at point `B` is `2` times the volume of `H_(3)A` used. Total volume of solution `= (100 mL + 200 mL) = 300 mL` `M_(1)V_(1) (Beofre) = M_(2)V_(2)(After)` `0.1 xx 100 = M_(2) xx 300` `M_(2) = (0.1)/(3) = (1)/(30) = 0.033` `:. [Na_(2)HA] = 0.033 M` `:. pH = (1)/(2) (pK_(w) + pK_(a_(3)) + log C)` `= (1)/(2) (14 + 5+ "log"(1)/(30)) = (1)/(2) (19 - 1.48) = 8.76` change is `pH = 9.7 - 8.76 = 0.94` |
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| 1452. |
In equalitative analysis, cations of graph `II` as well as group `IV` both are precipitated in the form of sulphides. Due to low value of `K_(sp)` of group `II` sulphides, group reagent is `H_(2)S` in the presence of dil. `HC1`, and due to high value of `K_(sp)` of group `IV` sulphides, group reagent is `H_(2)S` in the presence of `NH_(4)OH` and `NH_(4)C1`. In a solution containing `0.1M` each of `Sn^(2+), Cd^(2+)`, and `Ni^(2+)` ions, `H_(2)S`gas is passed. `K_(sp) of SnS = 8 xx 10^(-29), K_(sp) of CdS = 15 10^(-28), K_(sp) of NiS - 3 xx 10^(-21), K_(1) of H_(2)S = 1 xx 10^(-7), K_(2) of H_(2)S = 1 xx 10^(-14)` If `H_(2)S` is passed into the above mixture in the presence of `HC1`, which ion will be precipitated first?A. `SnS`B. `CdS`C. `NiS`D. `SnS` and `CdS` (both together) |
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Answer» Correct Answer - C `[S^(2-)]_(min)` for `SnS = (K_(sp)(SnS))/([Sn^(2+)])`, `[S^(2-)]_(min)` for `CdS = (K_(sp)(CdS))/([Cd^(2+)])`, `[S^(2-)]_(min)` for `NiS = (K_(sp)(NiS))/([Ni^(2+)])`, and `K_(sp) (NiS) lt K_(sp) (SnS) lt K_(sp) (CdS)` `:. NiS` will precipitate first. |
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| 1453. |
Which of the following combinations of solute would result in the formation of a buffer solution. a. `CH_(3)COOH +NaOH` in i. `1:1` mol ratio ii. `2:1` mol ratio iii. `1:2` mol ratio b. `NH_(4)CI=NH_(3)`in i. `1:1` mol ratio ii. `2:1` mol ratio iii. `1:2 `mol ratio c. `HCI +NaCI` d. `HCI +CH_(3)COOH` e. `NaH +HCI` |
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Answer» a. i. It will form salt of weak acid and strong base `(W_(A)//S_(B)) CH_(3)COONa`. ii. It will form acidic buffer, since `1 mol` of weak acid will be left unreacted and `1mol` of `CH_(3)COONa` will be formed. ii. `1mol` of `CH_(3) COONa` and `1mol` of `NaOH` is left. b. (i),(ii), and (iii) all are basic buffers since in all cases it is a mixtu8re of salt of `W_(A)//S_(A) (NH_(4)CI)` and `W_(B) (NH_(3))`. Moreover, `NH_(4)CI` and `NH_(3)` do not react. c. It is not buffer since it is a mixture of `S_(A)(HCI)` and salt of `S_(A)//S_(B) (NaCI)`. d. It is not buffer since it is a mixture of `S_(A) (HCI)` and `S_(B) (NaOH)`. |
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| 1454. |
How many of the following combinations of reactants will react less than `2%` of theroetically possible extent? a. `CH_(3)COOH +H_(2)O` b. `CH_(3)COO^(Theta) +H_(2)O` c. `CH_(3)COO^(Theta) +H_(3)O^(o+)` d. `CH_(3)COOH +KOH` e. `CH_(3)COOK +HCI (aq)` f. `HCI(g) +H_(2)O` g. `CI^(Theta)+H_(3)O^(o+)` h. `CI^(Theta) +H_(2)O` i. `overset(o+)NH_(4) + KOH` j. `overset(o+)NH_(4)+overset(Theta)OH` k. `NH_(3)+H_(2)O` l. `NH_(3)+H_(3)O^(o+)` m. `NH_(3) + HCI (aq)` n. `K^(o+) + overset(Theta)OH` |
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Answer» (a),(b),(g),(h) and (k) a. `CH_(3)COOH + H_(2)O rarr H_(3)O^(o+) +CH_(3)COO^(Theta) (lt 2%) W_(A)` b. `CH_(3)COO^(Theta) + H_(2)O rarr CH_(3)COOH +overset(Theta)OH (lt 2%)` Salt of `W_(A)` g. `CI^(Theta) + H_(3)O^(o+) rarr` No reaction h. `CI^(Theta) + H_(2)O rarr` No reaction k. `NH_(3) +H_(2)O rarr overset(o+)NH_(4) + overset(Theta)OH` n. `K^(o+) + overset(Theta)OH rarr` No reaction. |
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| 1455. |
The following equilibrium exists in aqueous solution, `CH_(3)COOH hArr CH_(3)COO^(-) + H^(+)` if dil. HCl is added, without change in temperature, theA. Concentration of `CH_(3)COO^(-)` will increaseB. Concentration of `CH_(3)COO^(-)` will decreaseC. The equilibrium constant will increaseD. The equilibrium constant will decrease |
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Answer» Correct Answer - B Due to common ion effect of `H^(+)`. |
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| 1456. |
`Ka_(1), Ka_(2)` and `Ka_(3)` are the respective constants for the following reactions `H_(2)S hArr H^(+) + HS^(-)` `HS^(-) hArr H^(+) + S^(2-)` `H_(2)S hArr 2H^(+) + S^(2-)` The correct relationship between `Ka_(1), Ka_(2)` and `Ka_(3)` isA. `K_(a_(3)) = K_(a_(1)) xx K_(a_(2))`B. `K_(a_(3)) = K_(a_(1)) + K_(a_(2))`C. `K_(a_(3)) = K_(a_(1)) - K_(a_(2))`D. `K_(a_(3)) = K_(a_(1)) //K_(a_(2))` |
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Answer» Correct Answer - A `K_(a_(3)) = K_(a_(1)) xx K_(a_(2))` |
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| 1457. |
How much moles of sodium propionate should be added to `1L` of an aqueous solution containing `0.020 mol` of propionic acid to obtain a buffer solution of `pH 4.75 `? What will be the `pH` if `0.010 mol` of `HCI` is dissolved in the above buffer solution. Compare the last `pH` value with the `pH of 0.010 M HCI` solution. Dissociation constant of propionic acid, `K_(a)`, at `25^(@)C` is `1.34 xx 10^(-5)`. |
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Answer» Correct Answer - B::C::D Let the moles of sodium propinote added be `a mol L^(-1)` For an acidic buffer. `pH = pK_(a) +"log" (["Salt"])/(["Acid"])` or `4.75 =- log (1.34 xx 10^(-5)) +"log" (a)/(0.020)` or `a = 1.5 xx 10^(-2)` `CH_(3)CH_(2)COOH hArr CH_(3)CH_(2)COO^(Theta) +H^(o+)` `CH_(3)CH_(2)COONa hArr CH_(3)CH_(2)COO^(Theta) +Na^(o+)` Rule A A A: When `HCI` is added, the amount of acid increases because propanote ions recat with hydrogen ions to form undissociated salt. Therefore, the amount of propanoate decreases. New `["Acid"] = 0.020 + 0.010 = 0.030` New `["Salt"] = 0.015 - 0.010 = 0.005` `pH = pK_(a) +"log" (["Salt"])/(["ACid"])` `=- log (1.34 xx 10^(-4)) +"log" (0.005)/(0.030) = 4.09` `pH` of `0.010 M HCI =- log (1xx 10^(-2)) = 2` pH of `HCI` is less than that of the buffer. |
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| 1458. |
A solution contains `0.1M H_(2)S` and `0.3M HCI`. Calculate the conc.of `S^(2-)` and `HS^(-)` ions in solution. Given `K_(a_(1))` and `K_(a_(2))` for `H_(2)S` are `10^(-7)` and `1.3xx10^(-7)` respectively. |
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Answer» `H_(2)S hArr H^(+)+HS^(-)` , `(K_(a_(1))=10^(-7))` `HS^(-) hArr H^(+) + S^(2-)` , `(K_(a_(2))=1.3xx10^(-13))` `HCI rarr H^(+)+CI^(-)` Due to common ion effect, the dissociation of `H_(2)S` is suppressed and the `[H^(+)]` in solution is due to HCI. `:. K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S])` `10^(-7)=([0.3][HS^(-)])/([0.1]) [.: [H^(+)]` form `HCI = 0.3]` `:. [HS^(-)]= (10^(-7)xx0.1)/(0.3)=3.3xx10^(-8)M` Further `K_(a_(1))=([H^(+)][S^(2-)])/([HS^(-)])` and `K_(a_(1))= ([H^(+)][HS^(-))]/([H_(2)S])` `:. K_(a_(1))xxK_(a_(2))= ([H^(+)]^(2)[S^(2-)])/([0.1])` `:. [S^(2-)]= (1.3xx10^(-20)xx0.1)/(0.09)` `=1.44xx10^(-20)M` |
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| 1459. |
The following equilibrium exists in an aqueous solution of hydrogen sulphide `H_(2)S hArr H^(+) + HS^(-)` If dilute HCl is added to an aqueous solution of `H_(2)S` without any change in temperatureA. The equilibrium constant will changeB. The concentration of `HS^(-)` will increaseC. The concentration of undissociated `H_(2)S` will decreaseD. The concentration of `HS^(-)` will decrease |
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Answer» Correct Answer - D In aqueous solution following euilibrium the dilute HCl `H_(2)S hArr H^(+) + HS^(-)` While adding the dilute HCl solution `(HCl hArr H^(+) + Cl^(-))` equilibrium is shift to the left side in `H_(2)S hArr H^(+) + HS^(-)`. |
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| 1460. |
What is `[HS^(-)] & [S^(2-)]` in solution of `0.01M HCl` and `0.1 M H_(2)S` ?(Given that for `H_(2)S,K_(a_(1))=10^(-7)and K_(a_(2))=10^(-14))` |
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Answer» `K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S])rArr 10^(-7)=10^(-7)=(0.01[HS^(-)])/(0.1)rArr [HS^(-)]=10^(-6)M.` `K_(a_(1))xxK_(a_(2))=([H^(+)]^(2)[S^(2-)])/([H_(2)S])` `[S^(2-)]=(10^(-21)xx0.1)/((0.01)^(2))=10^(-18)M` |
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| 1461. |
The dissociation constant of `H_(2)S` and `HS^(-)` are respectively `10^(-7)` and `10^(-13)` . The pH of 0.1 M aqueous solution of `H_(2)S` will beA. 4B. 3C. 5D. 2.5 |
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Answer» Correct Answer - A Since `K_(a2) lt lt lt K_(a1)` Thus, `H^(+)` ions are mainly obtained from the first step i.e., `H_(2)S hArr HS^(-) +H^(+)` `:. [H^(+)] =sqrt(KC)=sqrt(10^(-7) xx10^(-1))=10^(-4)` `:. pH =-log (10^(-4))=4` |
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| 1462. |
Ammonium carbamate decomposes as `NH_(2)COONH_(4)(s) hArr 2NH_(3)(g) +CO_(2)(g)` . The value of `K_(p)` for the reaction is `2.9 xx 10^(-5) atm^(3)`. If we start the reaction with 1 mole of the compound, the total pressure at equilibrium would beA. 0.0766 atmB. 0.0194 atmC. 0.194 atmD. 0.0582 atm |
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Answer» Correct Answer - D `NH_(2)COONH_(4)(s) hArR 2NH_(3)(g)+CO_(2)(g)` If partial pressure of `CO_(2)` at equilibrium is p Then partial pressure of `NH_(3)` is 2p `K_(p)=(p_(NH_(3)))^(2)xx(p_(CO_(2)))=(2p)^(2)(p)=4p^(3)` Now `4p^(3)=2.9 xx 10^(-5) `or `p=1.935xx10^(-2)` `=5.81 xx 10^(-2) atm ` |
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| 1463. |
When equal volumes of the following solutions are mixed, precipitation of `AgCl (K_(sp) = 1.8 xx 10^(-10))` will occur only withA. `10^(-4) M (Ag^(+))` and `10^(-4) M (Cl^(-))`B. `10^(-5) M (Ag^(+))` and `10^(-5) M(Cl^(-))`C. `10^(-6) M (Ag^(+))` and `10^(-6) M (Cl^(-))`D. `10^(-10) M (ag^(+))` and `10^(-10) M (Cl^(-))` |
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Answer» Correct Answer - A Ionic product `= [Ag^(+)][Cl^(-)] = (10^(-4))/(2) xx (10^(-4))/(2) = 2.5 xx 10^(-9) gt K_(sp)` (`because` equal volumes are mixed, so `[Ag^(+)] = (10^(-4))/(2)` and `[Cl^(-)] = (10^(-4))/(2)`) |
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| 1464. |
When equal volumes of following solution are mixed, precipitation of `AgCl` ? `(K_(sp)=1.8xx10^(-10))` will occur only withA. `10^(-4)M(Ag^(+))and 10^(-4)M(Cl^(-))`B. `10^(-5)M(Ag^(+))and 10^(-5)M(Cl^(-))`C. `10^(-6)M(Ag^(+))and 10^(-6)M(Cl^(-))`D. `10^(-10)M(Ag^(+))and 10^(-10)M(Cl^(-))` |
| Answer» Correct Answer - A::B | |
| 1465. |
The `K_(sp)` for `Al(OH)_(3)` is `2.0xx10^(-31)`. What is the value of `DeltaG^(@)` for the precipitation of `Al(OH)_(3)` of `25^(@)`C? `Al^(3+)(aq)+3OH^(-)(aq)rarrAl(OH)_(3)(s)`A. `-175KJmol^(-1)`B. `-14.7KJmol^(-1)`C. `14.7KJmol^(-1)`D. `175KJmol^(-1)` |
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Answer» Correct Answer - a |
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| 1466. |
The salt `Al(OH)_(3)` involved in the following two equilibria. `Al(OH)_(3)(s)hArrAl^(3+)(aq)+3OH^(-)(aq),K_(sp)` `Al(OH)_(3)(s)+OH^(-)(aq)hArrAl(OH)_(4)^(-)(aq),K_(c)` Which of the following relationship is CORRECT at which solubility is minimum?A. `[OH^(-)]=((K_(sp))/(K_(c)))^(1//3)`B. `[OH^(-)]=((K_(c))/(K_(sp)))^(1//3)`C. `[OH^(-)]=sqrt(((K_(sp))/(K_(c)))^(1//4))`D. `[OH^(-)]=((3K_(sp))/(K_(c)))^(1//4)` |
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Answer» Correct Answer - d |
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| 1467. |
The salt `Al(OH)_(3)` is involved in the following two equilibria, `Al(OH)_(3(s)) hArr Al_((aq))^(3+) + 3OH_((aq))^(o.), ksp` `Al(OH)_(3(s)) + OH_((aq))^(-) hArr [Al(OH)_(4)^(-)]_((aq)), kc` Which of the following relationship is correct at which solubility is minimum?A. `[OH^(-)] = ((Ksp)/(Kc))^(1//3)`B. `[OH^(-)] = ((Kc)/(Ksp))^(1//4)`C. `[OH^(-)] = sqrt(((Ksp)/(Kc))^(1//4))`D. `[OH^(-)] = ((3Ksp)/(Kc))^(1//4)` |
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Answer» Correct Answer - D `S= [Al_((aq))^(+3)] + [Al(OH)_(4(aq))]^(-)` `S = (Ksp)/([OH^(-)]^(3) + Kc[OH^(-)]` for minimum solubility `(ds)/(d(OH^(-))) = 0` `:. - (Ksp xx 3 )/([OH^(-)]^(4)) + Kc = 0,[OH^(-)] = ((3Ksp)/(Kc))^(1//4)` |
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| 1468. |
In the two gaseous equilibria (i) and (ii) , at a certain temperature, `(i) SO_(2)(g) +1//2 O_(2)(g) hArr SO_(3)(g) ....K_(1)` `(ii) 2SO_(3)(g) hArr 2SO_(2)(g) +O_(2)(g) .....K_(2)` If `K_(1) `is `4 xx 10^(-3)` then `K_(2)` will beA. `8.0 xx 10^(-3)`B. `16.0 xx 10^(-6)`C. `6.25 xx 10^(4)`D. `6.25 xx 10^(8)` |
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Answer» Correct Answer - C For the given reactions `K_(2)=(1)/((K_(1))^(2))` Thus `K_(2)=(1)/((4 xx 10^(-3))^(2))=6.25 xx 10^(4)` |
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| 1469. |
To prepare a buffer of pH `8.26`, amount of `(NH_(4))_(2)SO_(4)` to be added into 500mL of `0.01M NH_(4)OH` solution `[pK_(a)(NH_(4)^(+))=9.26]` is:A. 0.05 moleB. 0.025C. 0.10 moleD. 0.005 mole |
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Answer» Correct Answer - b |
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| 1470. |
The salt `ZN (OH)_(2)` is involved in the following two equilibria: `Zn(OH)_(2)(s) hArr Zn^(3+)(aq) +2 overset(Theta)OH(aq) K_(sp) = 1.2 xx 10^(-17)` `Zn(IH)_(2)(s)+2 overset(Theta)OH hArr Zn(OH)_(4)^(2-)(aq)K_(f) = 0.12` Calculate `|overset(Theta)OH|` at which solubility of `Zn(OH)_(2)` be a minimum. Also find the solubility of `Zn(OH)_(2)` at this `pH`. |
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Answer» a. `Zn(OH)_(2)(s) hArr Zn^(2+) +2 overset(Theta)OH` `K_(sp) = [Zn^(2+)] [overset(Theta)OH]^(2)` b. `Zn(OH)_(2)(s) + 2 overset(Theta)OH hArr (Zn(OH)_(4))^(2-)` `K_(f) = ([(Zn(OH)_(4))^(2-)])/([overset(Theta)OH]^(2))` Let the solubility be `S` is charge (a) and thus solubility will also be `S` in change (b). By coupling (a) and(b), `2Zn(OH)_(2)(s) hArr Zn^(2+) (aq) + (Zn(OH)_(4))^(2-)(aq)`. `K_(sp) xx K_(f) = S xx S` or `S = sqrt(K_(sp) xx K_(f)) = sqrt(2xx10^(-17)xx0.2) = 1.2 xx 10^(-9)M` Thus, total solubility will be `2S = 2 xx 1.2 xx 10^(-9) = 2.4 xx 10^(-9)M` This solubility is minimum when `[Zn^(2+)] [overset(Theta)OH]^(2) = 1.2 xx 10^(-17)` `1.2 xx 10^(-9) [overset(Theta)OH]^(2) = 1.2 xx 10^(-17)` `:. [overset(Theta)OH] = 10^(-4)M` Note: If only (a) change than solubility of `Zn(OH)_(2) =3sqrt((K_(sp))/(4)) =3sqrt((1.2xx10^(-17))/(4)) = 1.44 xx 10^(-6)M` and at this concentration `[overset(Theta)OH] = 2 xx 1.44 xx 10^(-6)M` `= 2.88 xx 10^(-6)M` |
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| 1471. |
A `500mL` of an equilibrium mixture of gaseous `N_(2)O_(4)` and `NO_(2)` at `25^(@)C` and `753mm` of `Hg` was allowed to react with enough water to make `250mL` of solution at `25^(@)C`. Assume that all the dissolved `N_(2)O_(4)` is converted to `NO_(2)` which disproportionates in water yielding a solution of nitrous acid and nitric acid. aAsume further that disproportionation reaction goes to completion and that none of the nitrous acid disproportionates. The equilibrium constant `(K_(p))` for `N_(2)O_(4)(g)hArr 2NO_(2)(g) 0.113` at `25^(@)C`. `K_(a)` for `HNO_(2)` is `4.5 xx 10^(-4) at 25^(@)C`. a. Write balanced equation for disproportionation. b. What is the molar concentration of `NO_(2)` and `pH` of the solution? c. What is osmotic pressure of solution? d. How many grams of lime `(CaO)` would be required to neutralise the solution? |
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Answer» a. `N_(2)O_(4) hAtt 2NO_(2)` Let `a mol` of `N_(2)O_(4)` and `bmol` of `NO_(2)` were present in equilibrium mixture. `:. (a+b) = (PV)/(RT) = (653 xx 0.5)/(760 xx 0.0821 xx 298)` `K_(p) = ((n_(NO_(2)))^(2))/((n_(N_(2)O_(4))))xx[(P)/((n_(NO_(2))+n_(N_(2)O_(4))))] ..(i)` `:. 0.113 = (b^(2))/(a) xx [(753)/(760 xx (a+b))]` `:. (b^(2))/(a(a+b)) = 0.114 ...(ii)` From equations, (i) and (ii), we get `(b^(2))/(a) = 0.114 xx 0.020 = 2.3 xx 10^(-3)` `b^(2) = 2.3 xx 10^(-3) a = 2.3 xx 10^(-3) xx (0.02 -b)` `b^(2) + 0.0023 b - 4.6 xx 10^(-5) = 0` `:. b = - (0.0023+-sqrt((0.0023)^(2)+4xx4.6xx10^(-5)xx1))/(2xx1)` `=(-0.0023+-sqrt(1.90 xx 10^(-4)))/(2)` `b = 5.73 xx 10^(-3)` `:.` By equaiton (i) `a = 0.014` b. `2NO_(2)(aq) + 2H_(2) (l) rarr NHO_(2)(aq) + H_(3)O^(o+)(aq) + NO_(3)^(Theta) (aq)` Total `NO_(2)` moles `=underset((From NO_(2)))(5.73xx10^(-3))+2xx` moles of `N_(2)O_(4)` `= 5.73 xx 10^(-3) + 2 xx 0.014` `= 0.0337` `{:(2NO_(2)(aq)+,2H_(2)O(l)rarr,HNO_(2)+,H_(3)O^(o+)(aq)+,NO_(3)^(o+)(aq)),(0.0337,,0,0,0),(0,,(0.0337)/(2),(0.0337)/(2),(0.0337)/(2)):}` `:. [HNO_(2)] = (0.337 xx 1000)/(2xx250) = 0.0674 M`, `[H^(o+)] = (0.337xx1000)/(2xx250) = 0.0674M`, Due to common ion effect `(H_(3)O^(o+)` furnished by `HNO_(3))`, the dissociation of `HNO_(2)` is suppressed. `K_(a) = ([H^(o+)][NO_(2)^(Theta)])/([HNO_(2)]) = (0.067 xx [NO_(2)^(Theta)])/(0.0674) = 4.5 xx 10^(-4)` `:. [NO_(2)^(Theta)] = 4.5 xx 10^(-4)` Also `pH =- log [H^(o+)] =- log 0.0674 = 1.17` c. `pi = iERT` (where `i =` Number of particles present in solution or it is Vant Hoff factor) `= 0.0674 xx 0.0821 xx 298 xx3` `(2NO_(2)` furnishes three particles) `= 4.95 atm` d. `Eq` of `CaO` required `= Eq of HNO_(2) + Eq` of `HNO_(3)` `= (0.0337)/(2) xx 1 + (0.0337)/(2) xx1 = 0.0337` `:.` Mole of `CaO` required `= (0.0337)/(2)or (W)/(56) = (0.0337)/(2)` `W_(CaO) = 0.924g` |
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| 1472. |
The following solutions are mixed: `500mL of 0.01 M AgNO_(3)` and `500mL` solution that was both `0.01M` in `NaCI` and `0.01M` in `NaBr`. Given `K_(sp) AGCI = 10^(-10), K_(sp) AgBr = 5 xx 10^(-13)`. Calculate the `[CI^(Theta)]` in the equilibrium solution.A. `5 xx 10^(-5)M`B. `2.5 xx 10^(-5)`C. `5 xx 10^(-3)M`D. `2.5 xx 10^(-3)M` |
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Answer» Correct Answer - C i. On dilution (equal volume) and if there were no precipitation. `[NO_(3)^(Theta)] = [Ag^(o+)] = [CI^(Theta)] = [Br^(Theta)] = (0.01)/(2) = 0.005 M = 5.0 xx 10^(-3) M`. `AgBr` is the more soluble salt (less `Ksp` means more soluble)and would take precendence in the precipitating reaction. ii. Assume `AgCI` does not precipitate. In this case `Ag^(o+)` and `Br^(Theta)` would be removed by precipitation and the concentration of these two ions in solution would remain eqial to each other. iii. `[Ag^(o+)] [Br^(Theta)] = sqrt(K_(sp) AgBr) = (5.0 xx 10^(-13))^((1)/(2)) = 7.1 xx 10^(-7)M`. iv. `Q_(sp)` or I.P. of `AgCI = [Ag^(o+)] [CI^(Theta)]` `= (7.1 xx 10^(-7)) (5.0 xx 10^(-3)M) = 3.5 xx 10^(-9)` `Q_(sp)` of `AgCI gt K_(sp)` of AgCI.(`because` Some `AgCI` most also precipitate) Hence the assumption in (ii) is wrong. v. Since both holides precipitate, it is a case of simultaneous solubilities. vi. By electronutrality `[Na^(o+)] + [Ag^(o+)] = [CI^(Theta)] + [Br^(Theta)] + [NO_(3)^(Theta)]` `0.01 + [Ag^(o+)] = [CI^(Theta)] +[Br^(Theta)] = 0.005` or `[CI^(Theta)] + [Br^(Theta)] - [Ag^(o+)] = 0.005 ...(1)` vii. `[Ag^(o+)] [CI^(Theta)] = 10^(-10) ...(2)` `[Ag^(o+)] [Br^(Theta)] = 5 xx 10^(-13).. (3)` viii. Divide (2) by (3), `([CI^(Theta)])/([Br^(Theta)]) = 200` This shows `Br^(Theta)` plays a significant role in the total anion concentration of the solution. xi. Moreover, `[Ag^(o+)]` must be negligible in (1) because of ionsolubility of two silver salts. x. Therefore assume in (1) that `[CI^(Theta)] = 0.005 = 5 xx 10^(-3)`. `[CI^(Theta)] = 5 xx 10^(-3)` xi. From (2), `[Ag^(o+)] = (10^(-10))/([CI^(Theta)]) = (10^(-10))/(0.005) = 2.0 xx 10^(-8)` |
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| 1473. |
Solubility of `AgBr` in water is `S_(1),` in `"0.01 M "CaBr_(2)` is `S_(2)` in `0.01 M NaBr` is `S_(3)` and in `0.05 M AgNO_(3)` is `S_(4)`. The correct order of these solubilities is `:`A. `S_(1)gt S_(2)gtS_(3)gtS_(4)`B. `S_(1)gt S_(2)=S_(3)gtS_(4)`C. `S_(4)gt S_(3)gtS_(2)gtS_(1)`D. `S_(1)gt S_(3)gtS_(2)gtS_(4)` |
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Answer» Correct Answer - 4 In water, no common ion effect will act hence `S_(1)` will be maximum . `{:([CaBr_(2)]=0.01,,,[Br^(-)]=0.02,,),([NaBr]=0.01,,,[Br^(-)]0.01,,),([AgNO_(3)]=0.05,,,[Ag^(+)]=0.05,,):}` More the concentration of common ion, less will be the solubility. `S_(1)gtS_(3)gtS_(2)gtS_(4)` |
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| 1474. |
Which of the following statements is correct for a solution saturated with `AgCl` and `AgBr` if their solubilities in moles per litre in separate solutions are `x` and `y` respectively?A. `[Ag^(+)]=[Br^(-)]xx[Cl^(-)]`B. `[Cl^(-)]gt[Br]`C. `[Br^(-)]lty`D. `[Ag^(+)]=x+y` |
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Answer» Correct Answer - b |
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| 1475. |
Which of the following statements is correct for a solution saturated with `AgCl` and `AgBr` if their solubilities in moles per litre in separate solutions are `x` and `y` respectively?A. `[Ag^(+)]=x+y`B. `[Ag^(+)]=[Br^(-)]+[Cl^(-)]`C. `[Br^(-)]=y`D. `[Cl^(-)]gtx` |
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Answer» Correct Answer - b |
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| 1476. |
Let the solubilities of `Agbr` in water and in `0.01M caBr_(2), 0.01M KBr`, and `0.05M AgNO_(3)`be `S_(1), S_(2),S_(3)` and `S_(4)`, respectively. Give the decreasing order of the solubilities. |
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Answer» i. `AgBr + H_(2)O rarr Ag^(o+) (aq) + Br^(Θ) (aq) …S_(1)` ii. `AgBr` in `0.01 M CaBr_(2) …S_(2)` `[Br^(Θ)]` added `= 0.01 xx 2 =0.02M` iii. `AgBr` in `0.01 M KBr …S_(3)` `[Br^(Θ)` added `= 0.01M` iv. `AgBr` in `0.05 M AgNO_(3) ..S_(4)` `[Ag^(o+)]` added `= 0.05M` Since both `Br^(Θ)` ions and `Ag^(o+)` ions act as common ions, so larger the concentration of `Br^(Θ)` or `Ag^(o+)` ion added, more is the superssion of ionisation of `AgBr` and hence less will the solubility of `AgBr`. Therefore, the decreasing solubility order. `S_(1) gt S_(3) gt S_(2) gt S_(4)` |
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| 1477. |
The `emf` of the following cell is observed to be `0.118V` at `25^(@)C`. `[Pt,H_(2)(1atm)|HA(100mL 0.1M||H^(o+)(0.1M)|H_(2)(1atm)|Pt]` a. If `30mL` of `0.2M NaOH` is added to the negative terminal of battery, find the emf of the cell. b. If `50mL` of `0.2 M NaOH` is added to the negative terminal of battery, find the emf of teh cell. |
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Answer» (a = anode, c = cathode) `E_(H_(2)) = - 0.059 (pH_(c) - pH_(a)) [[[H^(o+)]_(c)=,0.1M],[pH=1,]]` `0.118 =- 0.059 (1-pH_(a))` `- (0.118)/(0.059) = 1 - pH_(a)` `:. pH_(a) = 1+(0.118)/(0.059) = 3` `[H^(o+)]_(a) = 10^(-3)M` Since `HA` is a `W_(A)` `:. pH_(W_(A)) = (1)/(2) (pK_(a) - logC)` `3 = (1)/(2) (pK_(a) - log 0.1)` `:. pK_(a) = 5`. a. When `30mL` of `0.2M NaOH` `(=30 xx 0.2 = 6mmol)` is added, acidic buffer is formed. `{:(,HA +,NaOHrarr,NaA+,H_(2)O),("Initial",100xx0.1,6mmol,0,0),(,=10mmol,,,),("Final",10-6=4,0,6,-):}` `:. ["Salt"] = (6mmol)/(130mL)` `["Acid"]_("left") = 10 - 6 = (4mmol)/(130mL)` `:. pH_(a) = pK_(a) + log.((6//130)/(4//130))` `= 5 + log .(6)/(4) = 5 + log 3 - log2` `= 5 + 0.48 - 0.3 = 5.18` `E =- 0.059 (pH_(c) - pH_(a))` `=- 0.059 (1-5.18)` `= 0.246V` b. When `50mL` of `0.2M NaOH(=50 xx 0.2 = 10 mmol)` is added, salt of `W_(A)//S_(B)` is formed. `{:(,HA +,NaOHrarr,NaA+,H_(2)O),(Initial,rArr10mmol,10mmol,-,-),("Final",rArr0,0,10mmol,-):}` `:. ["Salt"] = 10//150 = 0.06M` `:. pH_(a) = (1)/(2) (pK_(w) + pK_(A) + logC)` `= (1)/(2) [(14 +5 + log (6 xx 10^(-2))]` `= (1)/(2) (19 + log 3 + log 2 - 2)` `= (1)/(2) (19 + 0.48 + 0.3 - 2) = 8.89` |
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| 1478. |
Calculate the simultaneous solubilities of `AgSCN` and `AgBr`. `K_(sp) (AgSCN) = 1.0 xx 10^(-12), K_(sp) (AgBr) = 5.0 xx 10^(-13)` |
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Answer» In this case, please note that the `K_(sp)` values of two slats are very similar. So the concentration of `Ag^(o+)` ions (the common ions) cannot be calculated from a single salt alone and we have to consider the equilibrium of the two slats simultaneously. Let the simulatenous solubilities of `AgSCN` and `AgBr` be `x` and `y`, respectively, in `molL^(-1)`. `{:(AgSCN(s)hArrAg^(o+)underset(x)(aq)+SCunderset(x)N^(Θ)(aq),,,,),(AgBr(s)hArrAg^(o+)underset(y)((aq))+Brunderset(y)((aq)),,,,):}` At equibrium: `[Ag^(o+)] = x +y, [SCN^(Θ)] = x, [Br^(Θ)] = y` `[Ag^(o+)] [Br^(Θ)] = K_(sp(AgBr))` and `[Ag^(o+)] [SCN^(Θ)] = K_(ap(AgSCN))` According to electircal change neutrality equation. Total positive change = Total negative change Note that: `[Ag^(o+)] = [Br^(Θ)] +[SCN^(Θ)]` [This is an Electrical change neturality equation] `[Ag^(o+)] = (K_(sp)(AgBr))/([Ag^(o+)]) + (K_(sp)(AgSCN))/([Ag^(o+)])` `rArr [Ag^(o+)] = sqrt(K_(sp(AgBr))+K_(sp(AgSCN)))` `rArr x +y = 1.22 xx 10^(-6) ....(i)` Also, `([Br^(Θ)])/([SCN^(Θ)]) = (y)/(x) = (K_(sp(AgBr)))/(K_(sp(AgSCN))) = 0.5 ...(ii)` Using equaitons (i) and (ii), we get `x = 8.0 xx 10^(-7)` |
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| 1479. |
Higher the amount of acid or base used to product a deinite change of pH in a buffer solution, higher will be its buffe capacity. Buffer capcity of solution is maximum under the following conditions [Salt] = [Acid](in acid buffer) [Salt] = [Base] (in base capacity] pH of buffer solution lies inthe range given below `pH = pH_(a) +-1` In other words,any buffer solution can be used as buffer up to two pH units only, depending upon the value of `pK_(a)` or `pK_(b)`. A buffer is said to be efficient when `pH = pK_(a)` or `pOH = pK_(b)` pH of `0.08 "mol" dm^(-3)` HOCl soluton is `2.85`. Calculate its ionization constant.A. `1.5 xx 10^(-5)`B. `2.5 xx 10^(-5)`C. `1.5 xx 10^(-8)`D. `2.25 xx 10^(-8)` |
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Answer» Correct Answer - B `[H^(+)] = sqrt(K_(a) xx C)` |
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| 1480. |
Higher the amount of acid or base used to product a deinite change of pH in a buffer solution, higher will be its buffe capacity. Buffer capcity of solution is maximum under the following conditions [Salt] = [Acid](in acid buffer) [Salt] = [Base] (in base capacity] pH of buffer solution lies inthe range given below `pH = pH_(a) +-1` In other words,any buffer solution can be used as buffer up to two pH units only, depending upon the value of `pK_(a)` or `pK_(b)`. A buffer is said to be efficient when `pH = pK_(a)` or `pOH = pK_(b)` The solubility product of `Al(OH)_(3)` is `2.7 xx 10^(-11)`. Calculate its solubility in `gL^(-1)` and also find out pH of this solution. (Atomic mass of `Al = 27u`).A. `11.47`B. `2.55`C. `10.6`D. `3.5` |
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Answer» Correct Answer - A `S = ((K_(sp))/(27))^(1//4)` |
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| 1481. |
Higher the amount of acid or base used to product a deinite change of pH in a buffer solution, higher will be its buffe capacity. Buffer capcity of solution is maximum under the following conditions [Salt] = [Acid](in acid buffer) [Salt] = [Base] (in base capacity] pH of buffer solution lies inthe range given below `pH = pH_(a) +-1` In other words,any buffer solution can be used as buffer up to two pH units only, depending upon the value of `pK_(a)` or `pK_(b)`. A buffer is said to be efficient when `pH = pK_(a)` or `pOH = pK_(b)` Calculate the volume of water requried to dissolve `0.1 g` lead (II) chloride to get a saturated solution . (`K_(sp)` of `PbCl_(2) = 3.2 xx 10^(-8)`, atomic mass of `Pb = 207 u`)A. `0.1798`B. `0.3`C. `0.652`D. `0.412` |
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Answer» Correct Answer - A `K_(sp) = [Pb^(+2)][Cl^(-)]^(2) =(S)` |
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| 1482. |
Higher the amount of acid or base used to product a deinite change of pH in a buffer solution, higher will be its buffe capacity. Buffer capcity of solution is maximum under the following conditions [Salt] = [Acid](in acid buffer) [Salt] = [Base] (in base capacity] pH of buffer solution lies inthe range given below `pH = pH_(a) +-1` In other words,any buffer solution can be used as buffer up to two pH units only, depending upon the value of `pK_(a)` or `pK_(b)`. A buffer is said to be efficient when `pH = pK_(a)` or `pOH = pK_(b)` A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution `(Q_(sp))` becomes greater than its solubility product. If the solubility of `BaSO_(4)` in water is `8 xx 10^(-4) "mol" dm^(-3)` calculate its solubility in `0.01 "mol" dm^(-3)` of `H_(2)SO_(4)`A. `6 xx 10^(-5)`B. `5 xx 10^(-3)`C. `2 xx 10^(-3)`D. `6 xx 10^(-8)` |
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Answer» Correct Answer - A `K_(sp) = 64 xx 10^(-8)` `K_(sp) = (S) (S + 0.01)` |
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| 1483. |
Higher the amount of acid or base used to product a deinite change of pH in a buffer solution, higher will be its buffe capacity. Buffer capcity of solution is maximum under the following conditions [Salt] = [Acid](in acid buffer) [Salt] = [Base] (in base capacity] pH of buffer solution lies inthe range given below `pH = pH_(a) +-1` In other words,any buffer solution can be used as buffer up to two pH units only, depending upon the value of `pK_(a)` or `pK_(b)`. A buffer is said to be efficient when `pH = pK_(a)` or `pOH = pK_(b)` Any buffer can be used as a buffer up to :A. `10pH` unitsB. `5pH` unitsC. `2pH` unitsD. `1pH` units |
| Answer» Correct Answer - C | |
| 1484. |
Which of the following statement(s) is (are) correct ?A. The solubility product is the product of concentration of ions of an electrolyte each raised to the power of its coefficient in the balanced chemical equation in a saturated solutionB. The solubility product of an electrolyte is a function of temperatureC. Cations of group III are precipitated as their hydroxides by `NH_(4)OH` in the presence of `NH_(4)CI` because the solubility products of these hydroxides are lowD. The ionic product with the concentration of an electrolyte |
| Answer» Correct Answer - A::B::C | |
| 1485. |
Which of the following statement(s) si (are) correct ?A. The pH of `1.0xx10^(-8)M` solution of HCI is 8B. The conjugate base of `H_(2)PO_(4)^(-)` is `HPO_(4)^(2-)`C. Autoprotolysis constant of water incerses with temperatureD. When a solution of aweak monoprotic acid is titrated against a strong base, at half-neutralization point `pH=(1//2)pK_(a)` |
| Answer» Correct Answer - B::C | |
| 1486. |
Which of the following is/are correct order(s) in terms of increaseing pH ?A. `NaOHgtCH_(3)COONagtNaClgtNH_(4)Cl`B. `NH_(4)ClgtNaClgtCH_(3)COONagtNaOH`C. `NaOHgtNaHCO_(3)gtKClgt(NH_(4))_(2)SO_(4)`D. `CH_(3)COOHgtCH_(3)COONagtNH_(4)Cl` |
| Answer» Correct Answer - A::C | |
| 1487. |
If the `pH` of `0.26 M HNO_(2)` is `2.5`, what will be its dissociation constant. |
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Answer» Correct Answer - A::C `C` is fairly high, then `(1-alpha) ~~1` So, `pH_(W_(A)) = (1)/(2) (pK_(a) - logC)` `2.5 xx2 = pK_(a) - log 0.26` `5 = pK_(a) + 0.585` `pK_(a) = 4.415` `K_(a) = "Antilog" (-4.415)` `= "Antilog" (-4-0.415 + 1-1)` `= "Antilog" (bar(5).585)` `= 3.8 xx 10^(-5)` |
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| 1488. |
`K_(1)` & `K_(2)` for oxalic acid are `6.5 xx 10^(-2)` and `6.1 xx 10^(-5)` respectively . What will be the `[OH^(-)]` in a `0.01 M` solution of sodium of sodium oxalateA. `9.6 xx 10^(-6)`B. `1.4 xx 10^(-1)`C. `1.2 xx 10^(-6)`D. `1.3 xx 10^(-8)` |
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Answer» Correct Answer - C The hydrolysis of `C_(2)O_(4^(2-))` is as follows `C_(2)O_(4^(2-)) + H_(2)O hArr HC_(2)O_(4)^(-) + OH^(-)` `[OH^(-)] = sqrt((K_(w) xx C)/(K_(2))) = sqrt((10^(-14) xx 10^(-2))/(6.1 xx 10^(-5)))` `= 1.2 xx 10^(-6)` |
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| 1489. |
If te solubility product of `BaSO_(4)` is `1.5 xx 10^(-9)` in water, its solubility in moles per litre, isA. `1.5 xx 10^(-9)`B. `3.9 xx 10^(-5)`C. `7.5 xx 10^(-5)`D. `1.5 xx 10^(-5)` |
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Answer» Correct Answer - B `BaSO_(4) hArr Ba^(2+) + SO_(4)^(--)` Solubility constant `= S xx S` `1.5 xx 10^(-19) = S^(2), S = sqrt(1.5 xx 10^(-19)), S = 3.9 xx 10^(-5)`. |
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| 1490. |
The solubility of `BaSO_(4)` in water is `2.33 g 100 mL^(-1)`. Calculate the percentage loss in weight when `0.2g` of `BaSo_(4)` is washed with a. `1L` of water b. `1L` of `0.01 NH_(2)SO_(4).[Mw_(BaSO_(4)) = 233 g mol^(-1)]` |
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Answer» a. Solubility is in general expressed in `gL^(-1)`, so soubility of `BaSo_(4) = 2.33 xx 10^(-3) gL^(-1)` Loss in weight of `BaSO_(4) =` Amount of `BaSO_(4)` soluble. `rArr %` loss `= (2.33 xx 10^(-3))/(0.2) xx 100 = 1.16%` b. `0.01 NH_(2)SO_(4) -= 0.01 N SO_(4)^(2-) ions` `-= 0.0055 M SO_(4)^(2-) ion (n` factor for `SO_(4)^(2-) = 2)` Now presence of `SO_(4)^(2-)` prior to washing `BaSO_(4)` will supress the solubility of `BaSO_(4)` (due to common ion effect. ) The supersion will be governed by `K_(sp)` value of `BaSO_(4)`. So, first calculate the `K_(sp)` of `BaSO_(4)`. Solubility of `BaSO_(4)` in fresh water `= 2.33 xx 10^(-3) gL^(-1)` `= (2.33 xx 10^(-3))/(33) ,mol L^(-1) = 10^(-5)M` `K_(sp)= [Ba^(2+)][SO_(4)^(2-)] = (10^(-5))^(2) = 10^(10)` Now let `x` be the solubility in `mol L^(-1) in H_(2)SO_(4)` `rArr [Ba^(2+)] `in solution `= x mol L^(-1)` and `[SO_(4)^(2-)]` in solution `= (x + 0.005) mol L^(-1)` Ionic product `= [Ba^(2+)] [SO_(4)^(2-)] = (x) (x + 0.005)` `K_(sp) =` Ionic product at equilibrium (saturation) `rArr 1.0 xx 10^(-10) = (x) (x + 0.005)` Assuming `x` to be a small number `(x + 0.005) ~~ 0.005` `rArr x = (10^(-10))/(0.005) = 2xx10^(-8)mol L^(-1)` `= 2 xx 10^(-8) xx 233 gL^(-1)` ` = 4.66 xx 10^(-6) gL^(-1)` `rArr 4.66 xx 10^(-6)g of BaSO_(4)` is washed away. `rArr % loss = (4.66 xx 10^(-6)xx100)/(0.2) = 2.33 xx 10^(-3) %` |
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| 1491. |
Any precipitate is formed whenA. Solution becomes saturatedB. The value of ionic product is less that than the value of solubility productC. The value of ionic product is equal than the value of solubility productD. The value of ionic product is greater than the value of solubility product |
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Answer» Correct Answer - D When the value of ionic roduct is greater than the value of solubility product, then precipitate is formed. |
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| 1492. |
The solubility of `BaSO_(4)` in water is `2.33 xx 10^(-3)` gm/litre. Its solubility product will be (molecular weight of `BaSO_(4) = 233`)A. `1 xx 10^(-5)`B. `1 xx 10^(-10)`C. `1 xx 10^(-15)`D. `1 xx 10^(-20)` |
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Answer» Correct Answer - B The solubility of `BaSO_(4)` in g/litre is given `2.33 xx 10^(-3)` `because` in mole/litre. `n = (W)/(m.wt) = 1 xx 10^(-5) = (2.33 xx 10^(-3))/(233)` Because `BaSO_(4)` is a compound `K_(sp) = S^(2) = [1 xx 10^(-5)]^(2) = 1 xx 10^(-10)`. |
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| 1493. |
A solution of `MgCl_(2)` in water has pHA. `lt 7`B. `gt 7`C. 7D. 14.2 |
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Answer» Correct Answer - A `MgCl_(2)+ 2H_(2)O hArr Mg(OH)_(2) + 2HCl` |
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| 1494. |
The pH of `10^(-7)M` NaOH isA. 7.01B. Between 7 and 8C. Between 9 and 10D. Greater than 10 |
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Answer» Correct Answer - B `10^(-7) M NaOH` means `[OH^(-)] = 10^(-7)`, pOH = 7 pH = 14 - 7 = 7. |
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| 1495. |
50 ml water is added to a 50 ml solution of `Ba(OH)_(2)` of strength 0.01 M. The pH value of the resulting solution will beA. 8B. 10C. 12D. 6 |
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Answer» Correct Answer - C `0.01 M Ba(OH)_(2) = 0.02 N (Ba(OH)_(2)` `N_(1)V_(1) = N_(2)V_(2)` `[0.02 N] xx [ 50 ml] = N_(2) xx 100 ml` `N_(2) = (0.02 xx 50)/(100) = 10^(-2) N, [OH^(-)] = 10^(-2) N` `pOH = 2` or pH = 12. |
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| 1496. |
Approximate pH of `0.01M` aqueous `H_(2)S` solution, when `K_(1)` and `K_(2)` for `H_(2)S` at `25^(@)C` are `1xx10^(-7)` and `1.3xx10^(-13)` respectively:A. 4B. 5C. 6D. 8 |
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Answer» Correct Answer - a |
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| 1497. |
The concentration of hydrogen ion in a sample of soft drink is `3.8 xx 10^(-3)M`. What is its `pH`? |
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Answer» `pH =- log [3.8 xx 10^(-3)] =- log (3.8) - log (10^(-3))` `=- 0.58 - (-3) =2.42` `pH = 2.42` (solution is acidic) |
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| 1498. |
A solution containing both `Zn^(2+)` and `Mn^(2+)` ions at a concentration of `0.01M` is saturated with `H_(2)S`. What is pH at which MnS will form a ppt ? Under these conditions what will be the concentration of `Zn^(2+)` ions remaining in the solution ? Given `K_(sp)` of `ZnS is 10^(-22)` and `K_(sp)` of MnS is `5.6 xx 10^(-16), K_(1) xx K_(2)` of `H_(2)S = 1.10 xx 10^(-21)`. |
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Answer» The minimum `[S^(2-)]` for the start of the precipitaitons is that which satisfies the `K_(sp)` of `MnS` is in which `[Mn^(2+)] = 0.01M` `[S^(2-)] = K_(sp) "of"(MnS)/([Mn^(2+)]) =(5.6 xx 10^(-16))/(0.01) = 5.6 xx 10^(-14)` The `[H^(o+)]` of the soln. having the above `[S^(2-)]` can be calculated form the `H_(s)S` dissociation eq. expression. `([H^(o+)]^(2)[S^(2-)])/([H_(2)S]) = ([H^(o+)]^(2)xx5.6xx10^(-14))/(0.01) =1.1 xx 10^(-21)` `:. [H^(o+)] = 4.43 xx 10^(-5)` and `pH = 4.35` If the `[H^(o+)]` is greater than `4.34 xx 10^(-5)M` than the `[S^(2-)]` will be less than `5.6 xx 10^(-14)M` and `MnS` will no longer ppt. from the soln. `:.` The conce of `Zn^(2+)` ion remaining in the soln. can be calculated form the `K_(sp)` of `ZnS`. `:. [Zn^(2+)] = (K_(sp) "of" ZnS)/([S^(2-)]) = (1.0 xx 10^(-22))/(5.6 xx 10^(-14)) = 1.76 xx 10^(-9)M` Thus, by peroply adjusting the `[H^(o+)]` in the soln. it is possible to precipitate effectively all of the `Zn` from the solution without precipitation any `Mn^(2+)` ion. |
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| 1499. |
The equilibrium constant `(K_(c))` for the reaction of a weak acid `HA` with strong base `NaOH`is `10` at `25^(@)C`. Which of the following are correct deductionA. The ionization constant `K_(a)` at `25^(@)C` is `10^(-5)`B. pH of a `0.01 M` aqueous solution of `HA `at `25^(@)C` will be `3.5`C. pH of a `0.01 M` aqueous solution of `NaA` at `25^(@)C` will be `9`.D. If `K_(b)` of weak base `BOH` is `10^(-4)` at `25^(@)C`, equilibrium constant for neutralization of HA with BOH at `25^(@)C` will be `10^(5)` |
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Answer» Correct Answer - A::B::C::D `underset(("Weak"))(HA) + NaOH hArr NaA + H_(2)O , K = 10^(9)` `K = 1/(K_(h)) hArr = 10^(-9)` A) `K_(h) = (K_(w))/(K_(a))` or `K_(a) = (K_(w))/(K_(h)) = 10^(-5)` B) `pH` of `0.01 M HA = 1/2 pK_(a) - 1/2 "log" C = 3.5` C) pH of `0.1 M aq.NaA = 7 + 1/2 pK_(a) + 1/2 "log" C = 9` `K = (1)/(K_(h)) = (K_(a) xx K_(b))/(K_(w)) = (10^(-5) xx 10^(-4))/(10^(-14)) = 10^(5)` |
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In the third group of qualitive analysis, the precipitating reagent is `NH_(4)CI//NH_(4)OH`. The function of `NH_(4)CI` is toA. increases the ionisation of `NH_(4)OH`.B. Supress the ionisation of `NH_(4)OH`.C. Convert the ions of group theird into their respective chlorides.D. Stabilise the hydroxides of group `III` cations. |
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Answer» Correct Answer - B Due to common ion `(NH_(4)^(o+))`, from `NH_(4)CI`, it supresses the ionisation of `NH_(4)OH`, so as to give less concentration, of `overset(Theta)OH` ions, so that only the hydroxides of `III` group can precipitate, not the hydroxdes of higher group, since `K_(sp)` of the hydroxides of `III` group is lower than the `K_(sp)` of the hydroxides of higher group. |
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