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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.
| 1301. |
Given the data at `25^(@)C`, `{:(AgI^(-) rarr AgI+e^(-),,E^(@) = 0.152 V),(Ag rarr Ag^(+) + e^(-),,E^(@) = -0.800 V):}` What is the value of `log K_(sp)` for AgI (2.303 RT/F = 0.059 V)A. `-8.12`B. `+8.612`C. `-37.83`D. `-16.13` |
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Answer» Correct Answer - D Applying `E_(I^(-)|AgI|Ag)^(@) = E_(Ag^(+)|Ag)^(@) + 0.059 log K_(sp) (AgI)` `log K_(sp) (AgI) = (E_(I^(-)|AgI|Ag)-E_(Ag^(+)|Ag))/(0.09) = (-0.152-0.8)/(0.059) = -16.13` |
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| 1302. |
Solubility product of a salt of AB is `1 xx 10^(-8) M ^(2)` in a solution in which the concentration of `A^(+)` ions is `10^(-3)M`. The salt will precipitate when the concentration of `B^(-)` ions is keptA. between `10^(-8)M` to `10^(-7)M`B. between `10^(-7) M` to `10^(-6)M`C. `gt 10^(-5) M`D. `lt 10^(-8) M` |
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Answer» Correct Answer - C `K_(sp)[AB]=[A^(+)][B^(-)] =10^(-8)` Putting `[A^(+)]=10^(-3)M` we get `[B^(-)]10^(-5)M` AB will precipitate out if ionic produt is more than solubility product. Hence, for the precipitate `[B^(-)]gt 10^(-5)M` |
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| 1303. |
Solubility product of a salt of AB is `1 xx 10^(-8) M ^(2)` in a solution in which the concentration of `A^(+)` ions is `10^(-3)M`. The salt will precipitate when the concentration of `B^(-)` ions is keptA. `10^(-6)M`B. `10^(-5)M`C. `10^(-4)M`D. `10^(-7)M` |
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Answer» Correct Answer - B `K_(sp)=1 xx 10^(-8)` `=[A^(+)][B^(-)]=[10^(-3)][B^(-)]` The salt will be precipitated if Ionic product `gt K_(sp)` `10^(-3) [B^(-)]gt 1 xx 10^(-8)` `[B^(-)]gt (10^(-8))/(10^(-3)) i.e., [B^(-)] gt 10^(-5) M`. |
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| 1304. |
The solubility product of a salt having general formula `MX_(2)` in water is : `4 xx 10^(-12)`. The concentration of `M^(2+)` ions in the aqueous solution of the salt isA. `2.0 xx 10^(-6) M`B. `1.0 xx 10^(-4) M`C. `1.6 xx 10^(-4) M`D. `4.0 xx 10^(-10) M` |
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Answer» Correct Answer - B `{:(MX_(2)hArr,M^(2+),+,2X^(-)),(,S,,2S):}` `rArr S = 3sqrt((K_(sp))/(4)) = 3sqrt((4 xx 10^(-12))/(4)) = 1.0 xx 10^(-4) M`. |
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| 1305. |
Assuming complete ionisation, the pH of 1.0 M HCl is 1. The molarity of `H_(2)SO_(4)` with the same pH isA. 0.2B. 0.1C. `2.0`D. 0.05 |
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Answer» Correct Answer - D pH = 1 means `[H^(+)] = 10^(-1) M` Hence `[H_(2)SO_(4)] = (10^(-1))/(2) = (1)/(20) =0.05 M`. |
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| 1306. |
The pH of a solution of `H_(2)SO_(4)` is 1. Assuming complete ionisation, find the molarity of `H_(2)SO_(4)` solution :A. 0.1B. 0.2C. 0.005D. `2.0` |
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Answer» Correct Answer - C Ph OF 0.1 HCl`=1` `H_(2)SO_(4) hArr H^(+)+HSO_(4)^(-)` `HSO_(4)^(-) hArr H^(+) +SO_(4)^(2-)` `:. H_(2)SO_(4)` ionises to give two `H^(+)` ions. `:. 0.5 M H_(2)SO_(4)` will give 0.1 M `H^(+)` ion `:. `the molarity of `H_(2)SO_(40 =0.05` |
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| 1307. |
Phenolphthalein does not act as an indicator for the titration betweenA. NaOH and `CH_(3)COOH`B. `H_(2)C_(2)O_(4)` and `KMnO_(4)` sol.C. `Ba(OH)_(2)` and HClD. KOH and `H_(2)SO_(4)` |
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Answer» Correct Answer - B Phenolphthalene indicator cannot be used for `H_(2)C_(2)O_(4)` and `KMnO_(4)` solution. |
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| 1308. |
The enthalpy of formation `(Delta H_(f))` of a compound from its elements isA. `Delta H_(f) lt0`B. `Delta H_(f)gt0`C. `Delta H_(f)=` infiniteD. `Delta H_(f)` may be `lt0` or `gt0` |
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Answer» Correct Answer - D The question are base on facts. |
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| 1309. |
If x mol `L^(-1)` is the solubility of `Kal(SO_(4))_(2)` , then `K_(sp)` is equal toA. `x^(3)`B. `4x^(4)`C. `x^(4)`D. `4x^(3)` |
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Answer» Correct Answer - B `KAl(SO_(4))_(2)hArr K^(+)+Al^(3+)+2SO_(4)^(2-)` If x is solubility in mol `L^(-1)`, then `[K^(+)]=x, [Al^(3+)]=x,[SO_(4)^(2-)]=2x` `K_(sp)=[K^(+)][Al^(3+)][SO_(4)^(2-)]=x xx x xx (2x)^(2)=4x^(4)` |
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| 1310. |
The solubility product of `Al_(2)(SO_(4))_(2)` is given by the expression.A. `[Al^(3+)][SO_(4)^(2-)]`B. `[Al^(3+)]^(2)[SO_(4)^(2-)]`C. `[Al^(3+)]^(3)[SO_(4)^(2-)]^(2)`D. `[Al^(3+)]^(2)[SO_(4)^(2-)]^(3)` |
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Answer» Correct Answer - D `K_(sp)=[Al^(+3)]^(2)[SO_(4)^(-2)]^(3)` |
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| 1311. |
Which of the following is not a Lewis acid ?A. `BF_(3)`B. `AlCl_(3)`C. `FeCl_(3)`D. `PH_(3)` |
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Answer» Correct Answer - D `BF_(3),AlCl_(3)` and `FeCl_(3)` are electron defficient compounds and as such are Lewis acid. `PH_(3)` is a Lewis base. |
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| 1312. |
Which species among the following is a Lewis acid but is not a Bronsted acid ?A. HClB. `BF_(3)`C. `NH_(2)`D. `O^(2-)` |
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Answer» Correct Answer - B `BF_(3)` can accept a pair of electrons but cannot yield a proton. |
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| 1313. |
The pH of `10^(-2) M Ca(OH)_(2)` isA. 12B. 12.3C. 11.3D. 2 |
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Answer» Correct Answer - B `[OH^(-)] = 2 xx 10^(-2)` `pOH = log(2 xx 10^(-2)) = 1.6990` `:. pH = 14 - 1.6990 = 12.3010`. |
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| 1314. |
pH of `10^(-8) M HNO_(3)` is (nearly)A. 7B. 6.96C. 7.2D. 6 |
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Answer» Correct Answer - B `{:(,HNO_(3),rarr,H^(+),+,NO_(3)^(-)),("Conc. before ionisation",10^(-8)M,,0,,0),("Conc. after ionisation",0,,10^(-8),,10^(-8)):}` `:. [H^(+)] = 10^(-8) m` but pH = 8 is not possible because it is acid. Now `[H^(+)] = 10^(-7) M` are already present in solution and since `10^(-8) lt 10^(-7)` and thus, it should not be neglected. `:. [H^(+)] = 10^(-8) + 10^(-7) = 10^(-7)(1.1) M` `= 1.1 xx 10^(-7)M` `:. pH = 6.9586`. |
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| 1315. |
The pH value of gasfric juice in humna stomach is about 1.8 and in intiestine , it is about 7.8. The `pk_(a)` value of aspirin is 3.5 Aspirin will beA. Ionised in the small intestine and almost unionised in the stomachB. Ionised in the stomach and almost unionised in the small intestineC. Completely ionised in the small intestine and in the stomach.D. Unionised in the small intestine and in the stomach. |
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Answer» Correct Answer - A `pK_(a)=3.5` `log K_(a)=3.5` `K_(a)=10^(-3.5)` Thus, `K_(a)` of aspirin is quite small. As such it is a weak acid. In the presence of a strong acid is gastric unionised. However, in the small intestine, when medium is slightly alkaline `(pH =7.8)` it will be ionised. |
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| 1316. |
pH of 0.1 M `NH_(3)` aqueous solution is `(K_(b) = 1.8 xx 10^(-5))`A. 11.13B. 12.5C. 13.42D. 11.55 |
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Answer» Correct Answer - A `NH_(4)OH hArr NH_(4)^(+) + OH^(-)` `K_(b) = C alpha^(2) , (1.8 xx 10^(-5))/(.1) = alpha^(2), alpha = 1.34 xx 10^(-3)` `[OH^(-)] = alpha. C = 1.34 xx 10^(-3) xx .1` `pOH = log 10(1)/(1.34 xx 10^(-4)) , pOH = 2.87` `pH = pOH = 14 , pH + 2.87 = 14` `pH = 14 - 2.87, pH = 11.13`. |
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| 1317. |
pH of human blood is 7.4. Then `H^(+)` concentration will beA. `4 xx 10^(-8)`B. `2 xx 10^(-8)`C. `4 xx 10^(-4)`D. `2 xx 10^(-4)` |
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Answer» Correct Answer - A `pH = - log[H^(+)], 7.4 = -log[H^(+)], [H^(+)] = 4 xx 10^(-8) M`. |
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| 1318. |
An alcoholic drink substance pH = 4.7 then OH ion concentration of this solution is `(K_(w) = 10^(-14) mol^(2)//l^(2))`A. `3 xx 10^(-10)`B. `5 xx 10^(-10)`C. `1 xx 10^(-10)`D. `5 xx 10^(-8)` |
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Answer» Correct Answer - B pH = 4.7 pH + pOH = 14, pH = 14- 4.7 , pOH = 9.3 `[OH^(-)] = "Antilog" [-9.3] [OH^(-)] = 5 xx 10^(-10)`. |
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| 1319. |
Aqueous solution of HCl has the pH = 4. Its molarity would beA. 4 MB. 0.4 MC. 0.0001 MD. 10 M |
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Answer» Correct Answer - C pH = 4 means, `[H^(+)] = 10^(-4)` mol/l. |
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| 1320. |
The dissociation constant of an acid HA is `1 xx 10^(-5)`. The pH of 0.1 molar solution of the acid will beA. FiveB. FourC. ThreeD. One |
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Answer» Correct Answer - C `HA hArr H^(+) + A^(-)` `[H^(+)] = 0.1 M , [H^(+)]^(2) = K_(a) xx C` `[H^(+)] = sqrt(K_(a) xx C) = sqrt(1 xx 10^(-5) xx 0.1) = sqrt(10^(-6))` `[H^(+)] = 10^(-3)M, pH = 3`. |
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| 1321. |
On addition of ammonium chloride to a solution of ammonium hydroxideA. Dissociation of `NH_(4)OH` increasesB. Concentration of `OH^(-)` increasesC. Concentration of `OH^(-)` decreasesD. Concentration of `NH_(4)^(+)` and `oH^(-)` increases |
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Answer» Correct Answer - C Due to common ion effect. |
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| 1322. |
An electrolyteA. Gives complex ions in solutionB. Dissolves in water to give ionsC. Is ionized in the solid stateD. Generates ions on passing electric current |
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Answer» Correct Answer - B Electrolytes are those substances which on dissolving in water give ions. |
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| 1323. |
Ammonium hydroxide is a weak base becauseA. It has low vapour pressureB. It is only slightly ionizedC. It is not a hydroxide of any metalD. It has low density |
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Answer» Correct Answer - B It is weak electrolyte since slightly ionized. |
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| 1324. |
In which of the following dissociation of `NH_(4)OH` will be minimumA. NaOHB. `H_(2)O`C. `NH_(4)Cl`D. NaCl |
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Answer» Correct Answer - C When we add `NH_(4)OH` in `NH_(4)Cl` solution ionization of `NH_(4)OH` is decreased due to common ion effect. |
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| 1325. |
Assertion (A): A solution contains `0.1M` each of `pB^(2+), Zn^(2+),Ni^(2+)`, ions. If `H_(2)S` is passed into this solution at `25^(@)C`. `Pb^(2+), Ni^(2+), Zn^(2+)` will get precpitated simultanously. Reason (R): `Pb^(2+)` and `Zn^(2+)` will get precipitated if the solution contains `0.1M HCI`. `[K_(1) H_(2)S = 10^(-7), K_(2)H_(2)S = 10^(-14), K_(sp) PbS =3xx 10^(-29) K_(sp) NiS = 3 xx 10^(-19). K_(sp) ZnS = 10^(-25)]`A. If both (A) and (R) are correc, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrect.D. If (A) is incorrect, but (R) is correct. |
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Answer» Correct Answer - A Here `0.1M H_(2)S` is passed into the solution directly. Selectively precipitation of `Pb^(2+), Zn^(2+)` and `Ni^(2+)` is not possible, as the ionic product to their respective sulphides is more than their `K_(sp)` values and `S^(-2)` ion is not added slowely or drowise. If `H_(2)S` is passed into the solution contianing only `H_(2)O`, then `[S^(2-)] ~~K_(2) of H_(2)S = 10^(-14)M`. But in `0.1M HCI` solution. `K_(1) xx K_(2) = ([H^(o+)]^(2)[S^(2-)])/(H_(2)S)` `10^(-7)xx 10^(-14) = ((0.1)[S^(2-)])/(0.1) :. [S^(2-)] = 10^(-20)M`. `:. Q_(sp)` of `PbS` and `ZnS(0.1 xx 10^(-20)) gt K_(sp) of PbS` and `Zns` (Hence both precipitate) `Q_(sp) of NiS (10^(-21)) lt K_(sp)` of `NiS` (Hence do not precipitate). |
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| 1326. |
If `Delta H_("ionisation")^(@)` for HCN and `CH_(3)COOH` are `45.2` and `2.1 KJ mol^(-1)` , which one of the following is correct ?A. `pK_(a(HCN))ltpK_(a(CH_(3)COOH))`B. `pK_(a(HCN))gtpK_(a(CH_(3)COOH))`C. `pK_(a(HCN))=pK_(a(CH_(3)COOH))`D. None of these |
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Answer» Correct Answer - B More is heat of ionisation, more is stability of molecule or lesser is dissociation. Thus `K_(a)(CH_(3)COOH)gtK_(a)(HCN)` Therefore `pK_(a)(HCN)gtpK_(a)(CH_(3)COOH)` |
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| 1327. |
The degree of dissociation of 0.1 M HCN solution is 0.01%. Its ionisation constant would beA. `10^(-3)`B. `10^(-5)`C. `10^(-7)`D. `10^(-9)` |
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Answer» Correct Answer - D `HA hArr H^(+) + A^(-)` `K = C alpha^(2) = 0.1 xx (10^(-4))^(2) = 10^(-9)`. |
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| 1328. |
Find out the `OH^(-)` ion concentration in `100 mL` of `0.015 M HCl` isA. `2.0 xx 10^(-9) M`B. `6.7 xx 10^(-13) M`C. `3 xx 10^(-10) M`D. `5 xx 10^(-12) M` |
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Answer» Correct Answer - B `[OH^(-)] = (Kw)/([H^(+)]` `= (1 xx 10^(-14))/(0.015)` `= 6.7 xx 10^(-13)` |
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| 1329. |
Find the `DeltapH(` initial `pH -` final `pH)` when `100 ml 0.01M HCl` is added in a solution containing `0.1m` moles of `NaHCO_(3)` solution of negligible volume `(k_(a_(1))=10^(-7),k_(a_(2))=10^(-11)` for `H_(2)CO_(3)) :`A. `6+2 log 3`B. `3-2 log 3`C. `3+2log2`D. `6-2log 3` |
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Answer» Correct Answer - 1 pH of `NaHCO_(3)` solution `=9` Now `H^(+)+HCO_(3)^(-)rarr H_(2)CO_(3)` `:.` no. of mmole of HCl remaining `=1-0.1=0.9` mmole `" ":. pH=-log(9xx10^(-3))=-2log3+3` |
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| 1330. |
A solution of `Na_(2)CO_(3)` is added drop by drop to 1 litre of a solution containing `10^(-4)` mole of `Ba^(2+)` and `10^(-5)` mole of `Ag^(+)`, if `K_(sp)` for `BaCO_(3)` is `8.1xx10^(-9)` and `K_(sp)` for `AgCO_(3)` is `6.9xx10^(-12)` then which is not true?A. No precipitate of `BaCO_(3)` will appear untill [`CO_(3)^(2-)`] reaches `8.1xx10^(-5)` mol per litre.B. A precipitate of `Ag_(2)CO_(3)` will appear when `[CO_(3)^(2-)]` reaches `6.9xx10^(-5)` mol per litre.C. No precipitate of `Ag_(2)CO_(3)` will appear untill `[CO_(3)^(-2)]` reaches `6.9xx10^(-2)` mole per litre.D. `BaCO_(3)` will be precipitate first. |
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Answer» Correct Answer - b |
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| 1331. |
A solution of `Na_(2)CO_(3)` is added drop by drop to litre of a solution containing `10^(-4)` mole of `Ba^(2+)` and `10^(-5)` mole of Ag, if `K_(SP)` for `BaCO_(3)` is `8.1xx10^(-9)` and `K_(SP)` for `Ag_(2)CO_(3)is 6.9xx10^(-12)`, then which is not true ?A. No precipitate of `BaCO_(3)` will appear untill `[CO_(3)^(2-)]` reaches `8.1xx10^(-5)` mole per litreB. A precipitate of `Ag_(2)CO_(3)` will appear when `[CO_(3)^(2-)]` reaches `6.9xx10^(-5)` mol `litre^(-1)`C. No precipitate of `Ag_(2)CO_(3)` will appear until `[CO_(3)^(2-)]` reaches `6.9xx10^(-2)` mol per litreD. `BaCO_(3)` will be precipitated first |
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Answer» Correct Answer - B For precipitation of `Ag_(2)CO_(3)`, `[CO_(3)^(2-)]= (K_(SP))/([Ag^(+)]^(2)) = (6.9xx10^(-12))/([10^(-5)]^(2))` `=6.9xx10^(-2)` and for precipitation of `BaCO_(3)`, `[CO_(3)^(2-)]= (K_(SP))/([Ba^(2+)])= (8.1xx10^(-9))/(10^(-4))=8.1xx10^(-5)` |
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| 1332. |
`pH` signifies:A. Puissance de hydrogenB. `-log [H^(o+)]`C. All the aboveD. `-14 -pOH` |
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Answer» Correct Answer - A `pH` stands for Fench word puissance de hydrogen which means power of hydrogen ions. |
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| 1333. |
Acetic acid and propionic acid have `K_(a)` values `1.75xx10^(-5)` and `1.3xx10^(-5)` respectively at a cetrain temperature. An equimolar solution of a mixture, of the two acids is partially neutralised by NaOH. How is the ratio of the contents of acetate and propionate ions related to the `K_(a)` values and the molarity ?A. `((alpha)/(1-alpha))=(1.75)/(1.3)xx((beta)/(1-beta))`, where `alpha` and `beta` are ionised fractions of these acidsB. The ratio is unrelated to the `K_(a)` valuesC. The ratio is unrelated to the molarityD. The ratio is unrelated to the pH of the solution |
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Answer» Correct Answer - A In a given mixture, the ionisatikon of two acids can be written as: Let `alpha, beta` be degree of ionisation at same concentration. `{:(CH_(3)COOHhArr,CH_(3)COO^(-)+,H^(+)),(1-alpha,alpha,alpha+beta):}` `{:(C_(2)H_(5)COOHhArr, C_(2)H_(5)COO^(-)+, H^(+)),(1-beta,beta,alpha+beta):}` `:. K_(A.A)= ([alpha][alpha+beta].c)/([1-alpha])` `K_(P.A)= ([beta][alpha+beta].c)/([1-beta])` `(K_(A.A))/(K_(P.A))=(alpha)/(1-alpha)xx(1-beta)/(beta)` or `(alpha)/(1-alpha)=(1.75)/(1.3)xx[(beta)/(1-beta)]` |
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| 1334. |
A certain weak acid has a dissociation constant `1.0xx10^(-4)`. The equilibrium constant for its reaction with a strong base is :A. `10 xx 10^(-4)`B. `10 xx 10^(-10)`C. `10 xx 10^(10)`D. `1.0 xx 10^(14)` |
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Answer» Correct Answer - C For the weak acid `HA`, `HA hArr H^(o+) + A^(Theta)` `K_(a) = ([H^(o+)]xx[A^(Theta)])/([HA])` Reaction of weak acid with strong base. `HA + OH^(Theta) hArr A^(Theta) + H_(2)O` `K = ([A^(Theta)])/([HA]xx[overset(Theta)OH])` Divident (1) by (2), we get `(K_(a))/(K) = [H^(o+)] [overset(Theta)OH] = K_(w)` or `K = (K_(a))/(K_(w)) = (10^(-14))/(10^(-14)) = 10^(10)` |
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| 1335. |
A certain weak acid has a dissociation constant `1.0xx10^(-4)`. The equilibrium constant for its reaction with a strong base is :A. `1xx10^(-4)`B. `1xx10^(-10)`C. `oo`D. `1xx10^(10)` |
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Answer» Correct Answer - D `HA hArr H^(+)+A^(-)` or `K_(a)=([H^(+)][A^(-)])/([HA])=10^(-4)" "....(i)` The reaction with strong base BOH can be expressed as `HA+OH^(-) hArr H_(2)O+A^(-)` `K=([A^(-)][H_(2)O])/([OH^(-)][HA])" "....(ii)` Also `K_(w)=[H^(+)][OH^(-)]=10^(-14)` From eqns. (i) , (ii) and (iii) `K=K_(a)//K_(w)=(10^(-4))/(10^(-14))=1010` |
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| 1336. |
When the ionic product of a solution exceeds the solubility product, the solution becomesA. saturatedB. unsaturatedC. a colloidD. supersaturated and precipitation of salt occurs. |
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Answer» Correct Answer - D When ionic product `gtK_(sp)` , the solution becomes supersaturated and thus precipitated of salt occurs. |
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| 1337. |
`H_(2)S` gas when passed through a solution of cations containing `HCl` precipitates the cations of second group in qualitative analysis but not those belonging to the fourth group. It is becauseA. Presence of HCl decreases the sulphide ion concentrationB. Sulphide of group II are unstable in HClC. Solubility product of group II sulphides is more than that of group IV sulphidesD. Presence of HCl increases the sulphide ion concentration. |
| Answer» Correct Answer - A | |
| 1338. |
`H_(2)S` gas when passed through a solution of cations containing `HCl` precipitates the cations of second group in qualitative analysis but not those belonging to the fourth group. It is becauseA. presence of HCl decreases the sulphide ion concentrationB. presence of HCl increases the sulphide ion concentrationC. solubility product of group II sulphides is more than that of group IV sulphidesD. sulphides of group IV cations are unstable in HCl |
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Answer» Correct Answer - A In qualitative analysis of cations of second group `H_(2)S` gas is passed in presence of HCl, therefore due to common ion effect, lower concentration of sulphide ions is obtained which is concentration of sulphide ions is obtained which is sufficient for the precipitation of second group cations in the form of their sulphides due to lower value of their solubility product `(K_(sp))`. Here, fourth group cations for precipitated because it require more sulphide ions for exceeding their ionic product to their solubility products which is not obtained here due to common ion effect. |
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| 1339. |
The solubility product of a sparingly soluble salt AB at room temperature is `1.21 xx 10^(-6)`. Its molar solubility isA. `1.21 xx 10^(-6)`B. `1.21 xx 10^(-3)`C. `1.1 xx 10^(-4)`D. `1.1 xx 10^(-3)` |
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Answer» Correct Answer - D AB is binary electrolyte. `S = sqrt(K_(sp)) = sqrt(1.2 xx 10^(-6)) = 1.1 xx 10^(-3) M`. |
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| 1340. |
If S and `K_(sp)` are respectively solubility and solubility product of a sparingly soluble binary electrolyte, thenA. `S = K_(sp)`B. `S = K_(sp)^(2)`C. `S = sqrt(K_(sp))`D. `S = (1)/(2) K_(sp)` |
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Answer» Correct Answer - C For binary electrolyte, so that `K_(sp) = S xx S = S^(2)` `S = sqrt(K_(sp))`. |
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| 1341. |
What is the `pH` at which `0.01MCo^(2+)` ions in solution precipiate down as `Co(OH)_(2)? K_(sp)` of `Co(OH)_(2)` is `2.5 xx 10^(-16)`. |
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Answer» `Co(OH)_(2) hArr Co^(2+) (aq) + 2OH^(Θ) (aq)` `[Co^(2+)] [overset(Θ)OH]^(2) = K_(sp) = 2.5 xx 10^(-16)` `(0.01) [overset(Θ)OH]^(2) = 2.5 xx 10^(-16)` `[overset(Θ)OH]^(2) = (2.5 xx 10^(-16))/(0.01) = 2.5 xx 10^(-14)` `[overset(Θ)OH] = sqrt(2.5 xx 10^(-14)) = 1.58 xx 10^(-7)M` `pOH =- log 1.58 xx 10^(-7) = 6.8` `pH + pOH = 14` and `pH = 14 - pOH = 14 - 6.8 = 7.2` |
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| 1342. |
The solubility product of `BaSO_(4)` at `25^(@)C` is `1.0 xx 10^(-9)`. What would be the concentration of `H_(2)SO_(4)` necessary to precipitate `BaSO_(4)` from a solution of 0.01 M `Ba^(2)` ionsA. `10^(-9)`B. `10^(-8)`C. `10^(-7)`D. `10^(-6)` |
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Answer» Correct Answer - C `BaSO_(4) hArr underset(underset(0.01)((S)))(Ba^(++)) + underset((S))(SO_(4)^(--))` `K_(sp) = S^(2) = S xx S = 0.01 xx S` `S_((SO_(4)^(2-))) = (K_(sp))/(S_((Ba^(++)))) = (1 xx 10^(-9))/(0.01) = 10^(-7)` mole/litre. |
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| 1343. |
Given that solubility product of `BaSO_(4)` is `1 xx 10^(-10)` will be precipiate from when a. Equal volumes of `2 xx 10^(-3)M BaC1_(2)` solution and `2 xx 10^(-4)M Na_(2)SO_(4)` solution, are mixed? b. Equal volumes of `2 xx 10^(-8) M BaC1_(2)` solution and `2 xx 10^(-3)M Na_(2)SO_(4)` solution, are mixed? c. `100mL` of `10^(-3)M BaC1_(2)`and `400mL` of `10^(-6)M Na_(2)SO_(4)` are mixed. |
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Answer» Correct Answer - A::B::C i. `BaC1_(2)` ionise completely in the solution as `BaC1_(2) rarr Ba^(2+) + 2C1^(Theta)` `[Ba^(2+)] = [BaC1_(2)] = 2 xx 10^(-3)M` (given) `Na_(2)SO_(4)` ionise completely in the solution as `Na_(2)SO_(4) rarr 2Na^(o+) + SO_(4)^(2-)` ` :. [SO_(4)^(2-)] = [Na_(2)SO_(4)] = 2 xx 10^(-4)M` (given) Since equal volume of the two solution are mixed together, the concentration of `Ba^(2+)` ions and `SO_(4)^(2-)` ions after mixing will be `[Ba^(2+)] = (2xx10^(-3))/(2)= 10^(-3)M` and `[SO_(4)^(2-)] = (2xx10^(-4))/(2) = 10^(-4)M` `:.` Ionic product of `BaSO_(4) = [Ba^(2+)] [SO_(4)^(2-)]` ` = 10^(-3) xx 10^(-4) = 10^(-7)` which is greater than the solubility product `(1 xx 10^(-10))` of `BaSO_(4)`. Hence, a precipitate of `BaSO_(4)` will be formed. ii. Here, the concentration before mixing is: `[Ba^(2+)] = [BaC1_(2)] = 2 xx 10^(-8)M` `[SO_(4)^(2-)] = [Na_(2)SO_(4)] = 2 xx 10^(-3)M` (Given) Concentration after mixing equal volumes will be `[Ba^(2+)] = (2 xx 10^(-8))/(2) = 10^(-8)M` `[SO_(4)^(2-)] = (2 xx 10^(-3))/(2) = 10^(-3)M` `:.` Ionic product of `BaSO_(4) = [Ba^(2+)] [SO_(4)^(2-)]` `= 10^(-8) xx 10^(-3) = 10^(-11)` whcih is less than the solubility product `(1 xx 10^(-10))`. Hence, no ppt will be formed in this case. iii. Concentration of `Ba^(2+)` (Total volume `= 100 + 400 = 500M`) `{:(M_(1)V_(1),=,M_(2)V_(2),,),(("Before"),,("After"),,):}` `100 xx 10^(-3) = M_(2 xx 500` `M_(2) = (100 xx 10^(-3))/(500) = (10^(-3))/(5)M` Concentration of `SO_(4)^(2-)` `{:(M_(1)V_(1),=,M_(2)V_(2),,),(("Before"),,("After"),,):}` `400 xx 10^(-6) = M_(2) xx 500` `M_(2) = (400 xx 10^(-6))/(500) = (4)/(5) xx 10^(-6)M` `IP (or) Q_(sp)` of `BasO_(4) = [Ba^(2+)] [SO_(4)^(2-)]` ` = (10^(-3))/(5) xx (4)/(5) xx 10^(-6)` `(4)/(25) xx 10^(-9) M^(-2)` `= 0.16 xx 10^(-9)M^(2) = 1.6 xx 10^(-10)M^(2)` `Q_(sp) gt K_(sp)` So precipitate of `BaSO_(4)` will taken there. |
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| 1344. |
Assuming that the only source of perodic group `IIA` metals is an equimolar mixture of `NaC1, BaC1_(2)` and `mGC1_(2)`, suggest ways of preparing pure samples of a. `MgSO_(4)` b. `Ba` metal c. `Ba (C_(2)H_(3)O_(2))_(2)`. |
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Answer» One possible method is given for each part (a) Add `NaOH`, precipiating `Mg(OH)_(2)`. The `Na^(o+)` and `Ba^(2+)` remain in solution. After filtering the `Mg(OH)_(2)`, add `H_(2)SO_(4)` to it, yielding `MgSO_(4)` in solution. Evaporate most of the water, causing `MgSO_(4)` to crystallise. (b) Add `Na_(2)CO_(3)` to the solution, remaining from part (a), precipitating `BaCO_(3)`. Add `HC1`. `BaCO_(3) + 2HC1rarr BaC1_(2) +H_(2)O + CO_(2)` Evaporate, melt the solid `BaC1_(2)`, and electrolyse. (c) To `BaCO_(3)` obtained in part (b), and `HC_(2)H_(3)O_(2)`, and evaporate. `2CH_(3)COOH rarr BaCO_(3) rarr (CH_(3)COO)_(2) Ba +H_(2)O +CO_(2)` |
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| 1345. |
How much of the following cations belong to group `IIA, III, IV`, and `V` only in quanlitative salt analysis? `S^(2+), Hg_(2)^(2+), Pb^(2+), Zn^(2+), Cu^(2+), Cr^(3+), Ag^(3+), Mg^(2+), Sb^(3+)` |
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Answer» Correct Answer - D III A group: `Pb^(2+), Cu^(2+)`. `III` group: `Cr^(3+)`, `IV` group: `Zn^(2+)`. |
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| 1346. |
Equilibrium constants of `T_(2)O` `(T or _(1)^(3)H is an isotope " of "_(1)^(1)H)` and `H_(2)O` are different at 298 K. Let at 298 K pure `T_(2)O` has pT (like pH) is 7.62. The pT of a solution prepared by adding 10 mL. of 0.2 M TCl to 15 mL of 0.25 M NaOT is:A. `2-"log"7`B. `14+"log"7`C. `13.24 - "log"7`D. `13.24 + "log"7` |
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Answer» Correct Answer - D `{:(TCl+NaOT ,rarr,NaCl+,T_(2)O),("Initial milli-moles 2",,,3.75),("milli-moles of remaining",,,NaOT = 1.75 "moles"):}` `NaOTcon. = (1.75)/(40)` `= 0.07 = 7 xx 10^(-2)` `pOT = 2 - log 7` `:. pT = 15.24 - 2 + "log" 7` `= 13.24 + "log" 7` |
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| 1347. |
For pure water `(pH=7), K_(w)` at 298 is `10^(-14)`. On adding some acid to it, its pH changes to 3. The value of `K_(w)` for this acidicfied water will beA. `10^(-11)`B. `10^(-14)`C. `10^(-6)`D. None of these |
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Answer» Correct Answer - B Factual question. |
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| 1348. |
At 298 K pure `T_(2)O` has `pT` (like pH) is `7.60`. Find out the pT of a solution prepared by adding `10 ml` of `0.2 M TCl` to `15 ml` of `0.20 M NaOT`. (Given: log 2 = 0.30) Multiply your answer with 10 to get answer. |
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Answer» Correct Answer - 138 |
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| 1349. |
For two different acids with same concentration:A. relative strenght is expressed as `(alpha_(1))/(alpha_(2))`B. relative strenght is expressed as `(K_(a_(1)))/(K_(a_(2)))`C. relative strenght is expressed as `sqrt((K_(a_(1))/(K_(a_(2))))`D. relative strenght is expressed as `(pH_(1))/(pH_(2))` |
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Answer» Correct Answer - a,c |
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| 1350. |
On adding an acid in a sample of pure water at `25^(@)C` :A. the ionic product of water becomes greater than `10^(14)M^(2)`B. the ionic product of water remains `10^(-14)M^(2)`C. `[H^(+)]` becomes greater than `[OH^(-)]`D. `[OH^(-)]` becomes zero. |
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Answer» Correct Answer - b,c |
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