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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.
| 501. |
Calculate the degree of ionisation of `0.05 M` acetic acid if its `pK_(a)` value is `4.74`. How is the degree of dissociation affected when its solution also contains a. `0.01 M`, b. `0.1 M` in `HCl`? |
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Answer» `{:(CH_(3)COOHhArr,CH_(3)COO^(-), +H^(+)), (1,0,0), (1-alpha,alpha,alpha):}` `(pK_(a)= - Iog K_(a)= 4.74, :. K_(a)= 1.82xx10^(-5))` `K_(a)= (C alpha^(2))/((1-alpha))=C alpha^(2)` , `(.: 1-alpha ~~1)` `:. alpha= sqrt((K_(a))/(C ))= sqrt((1.82xx10^(-5))/(0.05))` `= 0.019` or `1.9%` always calculate `alpha` first by `K_(b)= C alpha^(2)`, if `alphagt 5%` then use again `K= (C alpha^(2))/((1-alpha))` (a) If `H^(+)` are already present (due to HCI). `{:(CH_(3)COOHhArr,CH_(3)COO^(-), +H^(+)), (1,0,0.01), (C(1-alpha),Calpha,[0.01+Calpha]):}` `K_(b)= ([CH_(3)COO^(-)][H^(+)])/([CH_(3)COOH]) = (C alphaxx(0.01+C alpha))/(C(1-alpha))` Since presence of `H^(+)` will favour the reverse reaction or `alpha` will decrease, i.e., `0.01+Calpha=0.01` and `1-alpha=1` (due to common ion effect) `:. 1.82xx10^(-5) = (0.05xxalphaxx0.01)/(0.05)` `:. alpha=1.82xx10^(-3)= 0.0018` (b) Similary solve for `0.1 MHCL` `alpha= 0.00018` |
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| 502. |
A definite volume of an aqueous `N//20` acetic acid `(pK_(a) = 4.74)` is titrated with a strongs base. It is found that `75` equal-sized drops of `NaOH` added from a burette effect the complete neutralisation. Find the `pH` when an acid solution is neutralised to the extent of `20%, 40%`, and `80%`, respectively. |
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Answer» During the titration of weak acid with swtrong base, if weak acid is left bwhind, acidic buffer is formed. i. `20%` neutralistion of `W_(A)` means `20%` salf of `W_(A)//S_(B)` is formed and `80% W_(A)` is left. `:. pH = pK_(a) + "log"(["Salt"])/(["Acid")]` `= 4.74 +log ((20)/(80))` `=4.74 + log ((1)/(4))` `= 4.74 - log2^(2)` `= 4.74 - 2 xx 0.3 = 4.14` ii. `40%` neutralisation of `W_(A)` means `40%` salt of `W_(A)//S_(B)` is formed and `60% W_(A)` is left. `:. pH = 4.64 +log ((40)/(60))` `= 4.74 +log 2 - log3` `= 4.74 +0.3 - 0.48 = 4.56` iii. `90%` neutralisation of `W_(A)` means `90%` salt of `W_(A)//S_(B)` is formed and `10% W_(A)` is left. `:. pH = 4.74 + log ((90)/(10))` `= 4.74 + log 3^(2)` `= 4.74 +2 xx 0.48 = 5.70` |
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| 503. |
A definite volume of an aqueous `N//20` acetic acid `(pK_(a) = 4.74)` is titrated with a strongs base. It is found that `75` equal-sized drops of `NaOH` added from a burette effect the complete neutralisation. Find the `pH` when an acid solution is neutralised to the extent of `20%, 40%`, and `80%`, respectively.A. `4.14`B. `9.86`C. `5.34`D. `8.68` |
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Answer» Correct Answer - A `W_(A) (CH_(3)COOH)` is titrated with `S_(B)(NaOH)` and `20%` of the `W_(A)` is neutralised, so `20%` of the heat of `W_(A)//S_(B) (CH_(3)COONa)` is formed and `80%` of the `W_(A)(CH_(3)COOH)` is left. So an acidic buffer if formed. `:. ["Acid"] = 80, ["Salt"] = 20` `pH = pK_(a) + log [("Salt")/("Acid")]` `= 4.7447 + log [("Salt")/("Acid")]` `= 4.7447 + "log" (1)/(4)` `= 4.7447 -2 log2 = 4.7447 - 2xx 0.3010 = 4.14` |
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| 504. |
Which of the following is most soluble in water ?A. `MnS(K_(SP)=8xx10^(-37))`B. `ZnS(K_(SP)=7xx10^(-16))`C. `Bi_(2)S_(3)(K_(SP)=1xx10^(-70))`D. `Ag_(2)S(K_(SP)=6xx10^(-51))` |
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Answer» Correct Answer - B Find solubility for each separately by `S^(2)=K_(SP)` for MnS and ZnS. `108S^(5)=K_(SP)"for" Bi_(2)S_(3)` and `4S^(3)=K_(SP)"for" Ag_(2)S`. |
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| 505. |
The `pH` of an aqueous solution of `Ba(OH)_(2)` is `10`. If the `K_(sp)` of `Ba(OH)_(2)` is `1xx10^(-9)`, then the concentration of `Ba^(2+)` ions in the solution in `mol L^(-1)` isA. `1 xx 10^(-2)`B. `1 xx 10^(-4)`C. `1xx10^(-1)`D. `1xx10^(-5)` |
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Answer» Correct Answer - C `pH = 10, pOH = 4, [overset(Theta)OH] = 10^(-4)M`. `Ba(OH)_(2) rarr Ba^(2+) +2 overset(Theta)OH` `[overset(Theta)OH] = 10^(-4)`. (Since total concentration of `overset(Theta)Oh = 10^(-4)` `:. K_(sp) = [Ba^(2+)] [overset(Theta)OH]^(2)` `10^(-9) = x xx (10^(-4))^(2)` `:. x=10^(-1) =[Ba^(2+)]` |
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| 506. |
The `pH` of an aqueous solution of `Ba(OH)_(2)` is `10`. If the `K_(sp)` of `Ba(OH)_(2)` is `1xx10^(-9)`, then the concentration of `Ba^(2+)` ions in the solution in `mol L^(-1)` isA. `1xx10^(-2)`B. `1xx10^(-4)`C. `1xx10^(-1)`D. `1xx10^(-5)` |
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Answer» Correct Answer - C `pOH =14-10=4` `:. [OH^(-)] =1xx10^(-4) M` `K_(sp)=[Ba^(2+)][OH^(-)]^(2)` or `[Ba^(2+)]=K_(sp)//[OH^(-)]^(2)=(1xx10^(-9))/((1xx10^(-4))^(2))` |
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| 507. |
The pH of which compound in aqueous solution depends on its concentration in solution ?A. `underset(NH_(2))underset(|)(RCH).COOH`B. `NaHS`C. `CH_(3)COONH_(4)`D. `(NH_(4))_(2)SO_(4)` |
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Answer» Correct Answer - B pH of amphioprotic salt and weak acid + weak base salts is independent of concentration. |
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| 508. |
The degree of dissociation of `C_(6)H_(5)COOH` is influenced by :A. `NH_(4)OH`B. `NaOH`C. `HCI`D. either of these |
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Answer» Correct Answer - D Decreases due to common ion effect (HCI) or increases due to acid-base reaction. |
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| 509. |
If `K_(a)=10^(-5)` for a weak acid, then `pK_(b)` for its conjugate base would beA. `7`B. `5`C. `9`D. `6` |
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Answer» Correct Answer - C `Ka = 10^(-5) Ka xx Kb = Kw` |
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| 510. |
Assertion (A): If `HCI` gas is passed through satirated `NaCI` solution, solid `NaCI` starts separating out. `HCI` decrease the solubility product of `NaCI`.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 - C `(A)` is correct Due to common ion `(CI^(Theta))` effect, the supression of ionisation of `NaCI` occurs and `NaCI` starts precipitating. `(R)` is wrong. Correct `(R)` as in `(A)`. |
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| 511. |
How much `AgBr` could dissolve in `1.0L` of `0.4M NH_(3)`? Assume that `Ag(NH_(3))_(2)^(o+)` is the only complex formed. Given: the dissociation constant for `Ag(NH_(3))_(2)^(o+) hArr Ag^(o+) xx 2NH_(3)`, `K_(d) = 6.0 xx 10^(-8)` and `K_(sp)(AgBr) =5.0 xx 10^(-13)`. |
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Answer» Let solubility of `AgBr` be `xM`. Thus, `[Br^(Θ)] = xM`, but `[Ag^(o+)] != xM` since it will react with `NH_(3)` to form a complex and thus, its concentration will be decied by the disscoiation of complex. So, let `[Ag^(o+)] = yM`. `AgBr hArr Ag^(o+)(aq) +Br^(Θ) (aq)` `rArr K_(sp) = [Ag^(o+)] [Br^(-)] = yx = 5.0 xx 10^(-13)` Since the formation constant `(K_(f))` of the complex is very high, assume that whole of `Ag^(o+)` formed is consumed. `{:(Ag^(o+)+,2NH_(3)rarr,Ag(NH_(2))_(2).^(oplus),,),(x,0.4,,,),(,0.4-2x,x,,):}` `{:(Ag(NH_(3))_(2)^(o+)hArr,Ag^(o+)(aq),+2NH_(3),(K_(d)=5.0xx10^(-13)),,),(x,,0.4-2x,,),(x-y,,y0.4-2x+2y,,):}` Thus, `K_(d) = ([Ag^(oplus)][NH_(3)]^(2))/([Ag(NH_(3))_(2)^(oplus)])=(y(0.4-2x+2y)^(2))/(x-y) =6xx10^(-8)` Assuming `x-y ~~` since `K_(d)` is low and `x lt lt 0.4`, we get : `K_(d) = (y(0.4)^(2))/(x)` Solving for `x` : `x = 1.15 xx 10^(-3)M` (Verify the approximation yourself). |
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| 512. |
A buffer solution 0.04 M in `Na_(2)HPO_(4)` and 0.02 in `Na_(3)PO_(4)` is prepared. The electrolytic oxidation of 1.0 milli-mole of the organic compound RNHOH is carried out in 100 mL of the buffer. The reaction is RNHOH+H_(2)OrarrRNO_(2)+4H^(+)+4e^(-) The approximate pH of solution after the oxidation is complete is : `[Given : for H_(3)PO_(4),pK_(a1)=2.2,pK_(a2)=7.20,pK_(a3)=12]`A. `6.90`B. `7.20`C. `7.5`D. `8.2` |
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Answer» Correct Answer - C Hint: moles of `H^(+)` produced `= 4 xx 10^(-3)` `PO_(4)^(3-)(aq.) + H^(+)(aq.) rarr HPO_(4)^(2-)(aq.)` `{:("Initial",,0.002,0.004,,"__"),("milli moles",,,,,),("Final","___",0.002,,0.002,),("milli moles",,,,,):}` `HPO_(4)^(3-) + H^(+) (aq.) rarr H_(2)PO_(4)^(-) (aq.)` `{:("Initial milli",0.006,0.002,"___"),("moles",,,),("Final milli",0.004,"__",0.002),("moles",,,):}` finally buffr solution of `H_(2)PO_(4)^(-)` (acid) `HPO_(4)^(2-)` (conjugate base) is formed `pH =pK_(a_(2)) + "log" ([HPO_(4)^(2-)])/([H_(2)PO_(4)])` `= 7.2 + "log" (0.004)/(0.002) = 7.5` |
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| 513. |
Which statements is//are correct?A. `0.1M NH_(3)` solution will precipitate `Fe(OH)_(2)` from a `0.1M` solution `Fe^(2+)`.B. `0.1M NH_(3)` solution will not precipitate `Mg(OH)_(2)` from a solution which is `0.2M` in `overset(o+)NH_(4)` and `0.1M` in `Mg^(2+)`C. `0.1M NH_(3)` solution will not precipitate `AgOH` from a solution which is `0.01M`in `Ag^(o+)`.D. Will precipitate is part (c). |
| Answer» Correct Answer - A::B::C | |
| 514. |
What is `[Cd^(2+)]` in `1.0L` of solution prepared by dissolving `0.001mol Cd(NO_(3))_(2)` and `1.5 mmol. NH_(3) ? K_(d)` for the dissociation of `Cd(NH_(3))_(4)^(2+)` into `Cd^(2+)` and `4NH_(3)` is `1.8 xx 10^(-7)`. Neglect the amount of `Cd` in complexes containing fewer than `4` ammonia molecules. |
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Answer» `{:(,Cd^(2+)+,4NH_(3)hArr,Cd(NH_(3))_(4)^(2+),),("Initial",rArr0.001,1.5,0,),("Used up",rArr0.001-x,4(0.001-x),0.001-x,),("Eq.",rArr~~x,~~1.5,~~0.001,):}` `Cd(NH_(3))_(4)^(2+) hArr Cd^(2+) + 4NH_(3)`. `K_(d) = ([Cd^(2+)][NH_(3)]^(4))/([Cd(NH_(3))_(4)^(2+)])` `1.8 xx 10^(-7) = ((x)(1.5)^(4))/(0.001)` `:.x = 3.6 xx 10^(-16) = [Cd^(2+)]` |
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| 515. |
A buffer solution of `0.080M Na_(2)HPO_(4)` and `0.020 M Na_(3)PO_(4)` is prepared. The electrolytic oxidation of `1.0 mmol RNHOH` is carried out in `100mL` buffer to give `RNHOH + H_(2)O rarr RNO_(2) + 4H^(o+) + 4e^(-)` Calculate approximate `pH` of the solution after oxidation is complete `pK_(a_(2)), pK_(a_(2))`, and `pK_(a_(3))` of `H_(3)PO_(4)` are `2.12,7.20`, and `12.0`, respectively. |
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Answer» `[H^(o+)]` formed due to electrolyte oxidation `= 4mmol` `{:(,Na_(2)HPO_(4),+,Na_(3)PO_(4),),("Intial mmoles",0.08xx100,,0.02xx100,),(,=8,,=2,):}` `[H^(o+)]` will be used by `pO_(4)^(3-)` to give `H_(2)PO_(4)^(Theta)` `{:(PO_(4)^(3-)+,2H^(o+)rarr,H_(2)PO_(4)^(Theta),,),(2,4,0,,),(0,0,2,,):}` Thus, a new buffer containing `Na_(2)HPO_(4) (8mmol)` and `H_(2)PO_(4)^(Theta)` will remain in solution. `:. pH = pK_(a_(H_(2)PO_(4)^(Theta)) + "log"([Na_(2)HPO_(4)])/([H_(2)PO_(4)^(Theta)])` `= 7.2 + "log" (8)/(2) = 7.81` |
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| 516. |
When HCI gas is passed through a saturated solution of common salt, pure NaCI is Precipitated because:A. `HCI` is higly soluble in water.B. The ionic product `[Na^(o+)] [CI^(Theta)]` exceeds its solubility product `(K_(sp))`.C. The `K_(sp)` of `NaCI` is lowered the presence of `HCI^(Theta)` ions.D. `HCI` causes precipitation. |
| Answer» Correct Answer - A::B::D | |
| 517. |
Which statement is not true?A. `pH` of `1xx10^(8)M HCl` is `8`B. `96500` coulomb deposits 1 g equivalent of copperC. Conjugate base of `H_(2)PO_(4)^(-)` is `HPO_(4)^(2-)`D. `pH+pOH=14` for all aqueous solution |
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Answer» Correct Answer - A `pH` of `1xx10^(8)MHCI is = 7` (HCI is acid). |
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| 518. |
How much of the following are strong electorlytes? a. `NH_(3)` b. `NH_(4)CI` c. `CH_(3)COOH` d. `CH_(3)COONa` e. `HCI` f. `NaCI` |
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Answer» Correct Answer - D (b), (d),(e), and (f). All salts (b,d, and f) soluble hydroxides and the acid (e) are strong electorytes. |
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| 519. |
Which of the following pairs constitutes a bufferA. `NHO_(3)` and `NH_(4)NO_(3)`B. HCl and KClC. `NHO_(2)` and `NaNO_(2)`D. NaOH and NaCl |
| Answer» Correct Answer - A | |
| 520. |
Which of the following constitutes a buffer ?A. `HNO_(2)` and `NaNO_(2)`B. NaOH and NaClC. `HNO_(3)` and `NH_(4)NO_(3)`D. HCl and KCl |
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Answer» Correct Answer - A A pair constituent with `HNO_(2)` and `NaNO_(2)` because `HNO_(2)` is weak acid and `NaNO_(2)` is a salt of weak acid `(HNO_(2))` with strong base (NaOH). Hence, it is an example of acidic buffer solution. |
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| 521. |
How much of the following `0.1M` solutions are acidic? a. `NH_(4)CI` b. `NaOH` c. `HC_(2)H_(3)O_(2)` d. `NaCI` e. `NH_(3) +NH_(4)CI` f. `NH_(3)` g. `HCI` h. `HCIO_(4)` i.`(NH_(4))_(2)SO_(4)` j. `K_(2)SO_(4)` |
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Answer» Acidic `[HCIO_(4),HCI,HC_(2)H_(3)O_(2),(NH_(4))_(2)SO_(4) "and" (NH_(4)CI)/("Both salt of" W_(B)//S_(A))]` (Five compounds) |
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| 522. |
`0.6` mmol of NaCI and 1 mol of `HCI` in 1 L solution is a buffer. |
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Answer» `{:(NaCN+,HCIrarr,NaCI+,HCN),(0.1,1,0,0),(0,(1-0.5)=0.5,0.5,0.5):}` It becomes a mixture `HCN + NaCI + HCI` which is not a buffer. |
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| 523. |
Which of the following constitutes a buffer ?A. `HNO_(3)+NH_(4)NO_(3)`B. `HCI+KCI`C. `HNO_(2)+NaNO_(2)`D. `NaOH+NaCI` |
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Answer» Correct Answer - A The mixture contains weak acid `(HNO_(2))` and its conjugate base `NO_(2)^(-)`. |
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| 524. |
In which of the following combinations, is buffer action expected ? 1. `NH_(3)+NH_(4)CI` , 2. `HCI+NaCI` 3. `NH_(3)+HCI is 2:1` mole ration Select the correct answer using the code given below:A. 1 and 2B. 1 and 3C. 2 and 3D. 1,2 and 3 |
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Answer» Correct Answer - B `NH_(3)+HCI I n 2:1` will give `NH_(3)+NH_(4)CI I n 1:1` ratio. |
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| 525. |
Assertion (A): When small amount of acid or base is added to pure water, its `pH` undergoes a change. Reason (R) : Addition of an acid or a basic increases the degree of ionisation of water.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 - C `(A)` is correc. `(R)` is wrong. Correct `(R):` Addition of acid or base increases the `[H^(o+)]` or `[overset(Theta)OH]` and thus `pH` changes. |
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| 526. |
Given that `triangleH^(@) _(ionization )` of HX and HY are 30 and `25 kJ//mol` respectively, then which of the following relation `is // are correct?`A. `pK_(b)(X^(-))ltpK_(b)(Y^(-))`B. `pK_(a)(HX)gtpK_(a)(HY)`C. `pK_(a)(HX)ltpK_(a)(HY)`D. `pK_(b)(X^(-))gtpK_(b)(Y^(-))` |
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Answer» Correct Answer - a,b |
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| 527. |
A compound whose aqueous solution will have the highest `pH`A. NaClB. `NaHCO_(3)`C. `Na_(2)CO_(3)`D. `NH_(4)Cl` |
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Answer» Correct Answer - C The hydrolysis of NaCl gives neutral solution because it is salt of strong acid and strong base and hence, its pH is 7. `NH_(4)Cl` is salt of weak base and strong acid, so its pH is less than 7. `NaHCO_(3)` is also acidic whereas `Na_(2)CO_(3)` is salt of strong base and weak acid, so its pH is more than 7. |
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| 528. |
Aqueous solution of acetic acid containsA. `CH_(3)COO^(-)` and `H^(+)`B. `CH_(3)COO^(-),H_(3)O^(+)`C. `CH_(3)COO^(-),H_(3)O^(+)` and `H^(+)`D. `CH_(3)COOH,CH_(3)COO^(-)` and `H^(+)` |
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Answer» Correct Answer - d The aqueous solution of acetic acid ionise as follows : `H_(2)O+underset("Base")(CH_(3)COOH)hArr underset("Acid")(CH_(3)COO^(-))+H_(3)O^(+)` So, the aqueous solution of acetic acid contains `CH_(3)COO^(-), H_(3)O^(+)` and `CH_(3)COOH`. |
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| 529. |
`0.1M` solution of which of the substances will behave basic?A. Sodium borateB. Cacium nitrateC. `NH_(4)Cl`D. Sodium sulphate |
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Answer» Correct Answer - A On hydrolysis sodium horate form sodium hydroxide and boric, so the solution will show basic character because sodium hydroxide is strong base and boric acid is weak acid. While solution of sodium sulphate is netural and that of `NH_(4)Cl` and calcium nitrate is acidic. |
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| 530. |
At infinite dilution, the percentage dissociation of both weak acid and weak base is:A. `1%`B. `20%`C. `50%`D. `100%` |
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Answer» Correct Answer - D At infinite dilution, weak electrolytes are `100%` dissociated. |
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| 531. |
`20 mL` of `0.1N HCI` is mixed with `20 ml` of `0.1N KOH`. The `pH` of the solution would beA. `0`B. `7`C. `2`D. `9` |
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Answer» Correct Answer - B `HCI = 20 xx 0.1 = 2 mEq` `KOH = 20 xx0.1 = 2 mEq` `2 mEq HCI` combines with `2 mEq` of `KOH` and form `KCI`, a salt of `S_(A)//S_(B)` which do not hydrolyses and given neutral solution with `pH = 7`. |
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| 532. |
`0.1M` solution of which of the substances will behave basic?A. Sodium borateB. Ammonium ditorideC. Calcium nitrateD. Sodium sulphate |
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Answer» Correct Answer - A Salt of `S_(B)//W_(A)` will be basic. a. Sodium borate, salt of `S_(B)//W_(A) (NaOH +H_(3)BO_(3))`. It given basic solution. b. `NH_(4)CI`, salt of `w_(B)//S_(A), (NH_(4)OH +HCI)`. It hydrolyses and given acidic solution. c. `Ca(NO_(3))_(2)`, salt of `w_(B)//S_(A)`. `[Ca(OH)_(2)+HNO_(3))`, and gives acidic solution. d. `Na_(2)SO_(4)`, salt of `S_(A)//S_(B)`, which do not hydrolyses, gives neutral solution with `pH =7`. |
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| 533. |
Two sparnigly soluble salts `AX` and `BX_(2)` have their solubility product constant equal . Which of the following is (are) correct deduction(s) ?A. Solubility of `AX` isw greater than solubility of `BX_(2)`.B. If `S_(1)` and `S_(2)` are molar solubility of `AX` and `BX_(2)` then `S_(1) = (S_(2))^(3//2)`C. If X is a conjugate base of a weak acid, addition of `HNO_(3)` will increase solubility of both AX with `BX_(2)`.D. Increasing the temperature, increase the solublity of both `AX` and `BX_(2)`. |
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Answer» Correct Answer - A::C `K_(sp) = (AX) = S_(1)^(2)` `K_(sp) = (BX)_(2) = 4S_(2)^(3)` If `S_(1)^(2) = 4S_(2)^(3)`, then `S_(1) = sqrt(4S_(2)^(3)) = 2(S_(2))^(3//2)` `S_(1) gt S_(2)` |
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| 534. |
Strong acids are generally used as standard solution in acid-base titrations because:A. the pH at equivalence point will be 7B. they titrate both strong and weak baseC. they form more stable solutions than weak acidsD. the salts of strong do not hydrolyse |
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Answer» Correct Answer - B It is a fact. |
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| 535. |
For the reaction in aqueous solution `Zn^(2+) + X^(-) hArr ZnX^(+)`, the `K_(eq)` is greatest when X isA. `F^(-)`B. `NO_(3)^(-)`C. `ClO_(4)^(-)`D. `I^(-)` |
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Answer» Correct Answer - A Because of `F^(-)` is a highly electronegative. So it is easily lose the electron and reaction occur rapidly. |
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| 536. |
Explain why `0.1M NH_(3)` solution: a. Will precipitate `Fe(OH)_(2)` from `0.1M` soluton `Fe^(2+)`. b. Will not precipitate `Mg(OH)_(2)` from a solution which is `0.2 M` in `overset(o+)NH_(4)` and `0.01M` in `Mg^(2+)`. c. Will not precipitate `AgOH` from a solution which is `0.01M` in `Ag^(o+)`. |
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Answer» a. `NH_(3)` provides sufficient `overset(Θ)OH` to excess `K_(sp)` of `Fe(OH)_(2)`. b. `overset(o+)NH_(4)` makes a buffer solution, which limits `[overset(Θ)OH]` to such a level insufficient to cause precipitation of `Mg(OH)_(2)`. c. `NH_(3)` forms `Ag(NH_(3))_(2)^(o+)` complex ion, which is soluble. |
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| 537. |
A solution is 0.10 M in `Ag^(+), Ca^(2+), Mg^(2+)` and `Al^(3+)` ions. Which compound will precipitate at the lowest `[PO_(4)^(3-)]` when a solution of `Na_(3)PO_(4)` is added ?A. `Ag_(3)PO_(4) (K_(sp)=1xx10^(-16))`B. `Ca_(3)(PO_(4))_(2) (K_(sp)=1xx10^(-33))`C. `Mg_(3)(PO_(4))_(2) (K_(sp)=1xx10^(-24))`D. `AlPO_(4) (K_(sp)=1xx10^(-20))` |
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Answer» Correct Answer - d |
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| 538. |
Why are strong acids genetally used as standard solutions in acid-base titrationsA. The pH at the equivalence point will always by 7B. They can be used to titrate both strong and weak basesC. Strong acids form more stable solutions than weak acidsD. The salts of strong acids do not hydrolysed |
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Answer» Correct Answer - B Strong acid can be used to titrate both strong and weak base. |
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| 539. |
Given that `K_(a)` for acetic acid as `1.8 xx 10^(-5)` and `K_(b)` for `NH_(4)OH` As `1.8 xx 10^(-5)` AT `25^(@)C`, predict the nature of aqueous solution of ammonium acetateA. acidicB. basicC. Slightly acidic or basicD. Neutral |
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Answer» Correct Answer - D When `K_(a)=K_(b)` , the solution is neutral. |
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| 540. |
Out of `Ca^(2+) ,Al^(3+), Fe^(3+), Mg^(2+)` and `Zn^(2+)` the reagents `NH_(4)Cl` and aqueous `NH_(3)` and precipitateA. `Ca^(2+), Al^(3+)`B. `Al^(3+), Fe^(3+)`C. `Fe^(3+), Mg^(2+)`D. `Mg^(2+), Zn^(2+)` |
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Answer» Correct Answer - B Low solubility product |
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| 541. |
Solution of `0.1 N NH_(4)OH` and `0.1 N NH_(4)Cl` has `pH 9.25`, then find out `K_(b)` of `NH_(4)OH`.A. 9.25B. 4.75C. 3.75D. 8.25 |
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Answer» Correct Answer - B As `[NH_(4)Cl]=[NH_(4)OH],pOH=pK_(b)`. `:. pK_(b)=pOH =14-pH=14-9.25=4.75` |
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| 542. |
An environment chemist needs a carbonate buffer of `pH=10` to study the effects of the acidification of limestones rich solis.How many grams of `Na_(2)CO_(3)` must be added to `1.5L` of freshly prepared `0.2M NaHCO_(3)` to make the buffer.? For `H_(2)CO_(3),K_(a_(1))=4.7xx10^(-7),K_(a_(2))=4.7xx10^(-11)` |
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Answer» `pK_(a)=11-log4.7` `pK_(a)=11-log4.7` `pH=pK_(a)+log((n_(Na_(2)CO_(3)))/(0.3)):.10=11-log4.7+log((n_(Na_(2)CO_(3)))/(0.3))` `log((4.7xx0.3)/(n_(Na_(2)CO_(3))))=1rArr n_(Na_(2)CO_(3))=0.141` `W_(Na_(2)CO_(3))=0.141xx106=14.946~~15 gram`. |
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| 543. |
The strongest base from the following species isA. `NH^(2-)`B. `OH^(-)`C. `O^(2-)`D. `S^(2-)` |
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Answer» Correct Answer - A `NH_(4)^(+)` is the weakest acid. So its conjugate base is strongest. |
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| 544. |
Which of the following compounds are diproticA. `H_(2)PO_(5)`B. `H_(2)S`C. `HClO_(3)`D. `H_(3)PO_(3)` |
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Answer» Correct Answer - B::D Diprotic solvents give `2H^(+)` ions or `OH^(-)` ions. |
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| 545. |
Aqueous solution of `NH_(4)Cl` is …… in nature due to behaviour of….ion in solution:A. acidic,`NH_(4)`B. alkalline ,`NH_(4)`C. acidic,`Cl^(-)`D. alkalline,`Cl^(-)` |
| Answer» `NH_(4)Cl+H_(2)OhArrNH_(4)OH+Cl^(-)+H^(+)("Acidic solution due to hydrolysis of" NH^(4+)` ion). | |
| 546. |
Calculate the pH of `0.5L` of a`0.2 M NH_(4)Cl-0.2 M NH_(3)` buffer before and after addition of (a)` 0.05` mole of `NaOH` and (b) `0.05` mole of `HCl`. Assume that the volume remains constant.[Given:`pK_(a)`of `NH_(3)=4.74]` |
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Answer» `pOH=4.74+log((0.1)/(0.1))=4.74` `pH=9.26` (a) `{:(,NH_(4)^(+),+,H^(+),rarr,NH_(4)^(+)),(t=0,0.1"mole",,0.05"mole",,0.1"mole"),(,0.05"mole",,-,,0.15"mole"):}` `pOH=4.74+log((pOH=4.74+log((0.15)/(0.05))=4.26` `:. pH=9.74` (b)` {:(,NH_(3)^(+),+,H^(+),rarr,NH_(4)^(+)),(t=0,0.1"mole",,0.05"mole",,0.1"mole"),(,0.05"mole",,-,,0.15"mole"):}` `pOH=4.74+log((0.15)/(0.05))=5.22` `:.pH=8.78`. |
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| 547. |
Amines behave asA. Aprotic acidB. Neutral compoundC. Lewis acidD. Lewis base |
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Answer» Correct Answer - D `R - NH_(2)` (Amines) behaves as a Lewis base because it is capable of donating a lone pair of electron. |
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| 548. |
`K_(sp)` for `Cr(OH)_(3)` is `2.7 xx 10^(-31)`. What is its solubility in moles/litreA. `1 xx 10^(-8)`B. `8 xx 10^(-8)`C. `1.1 xx 10^(-8)`D. `0.18 xx 10^(-8)` |
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Answer» Correct Answer - A `{:(Cr(OH)_(3),rarr,Cr^(+3)+,3OH^(-)),(,,x,3x):}` `K_(sp) = x.(3x)^(3) = 27 x^(4)` `x = 4sqrt((K_(sp))/(27)) , x = 4sqrt((2.7 xx 10^(-31))/(27))` `x = 1 xx 10^(-8)` mole/litre. |
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| 549. |
According to Bronsted-Lowry concept, the correct order of relative strength of bases follows the orderA. `CH_(3)COO^(-) gt Cl^(-) lt OH^(-)`B. `CH_(3)COO^(-) gt OH^(-) gt Cl^(-)`C. `OH^(-) gt CH_(3)COO^(-) gt Cl^(-)`D. `OH^(-) gt Cl^(-) gt CH_(3)COO^(-)` |
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Answer» Correct Answer - C Relative of bases can be shown by their conjugate acids. Conjugate acid or `OH^(-)` is `H_(2)O` which is a weak acid conjugate acid of `CH_(3)COO^(-)` is `CH_(3)COOH` which is stronger than `H_(2)O`. While conjugate acid of `Cl^(-)` is HCl which is strongest out of there. so the order of relative strength of bases is `OH^(-) gt CH_(3)COO^(-) gt Cl^(-)`. |
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| 550. |
Calculate the pH of resulting solution, when 50 mL of `0.20M HCI` is mixed with 50 mL of `0.20M CH_(3)COOH`. |
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Answer» After mixing, volume becomes 100 mL. `:. M_(HCI)=("milli.mole")/("Total volume")=(50xx0.2)/(100)=0.1` `M_(CH_(3)COOH)=(50xx0.2)/(100)=0.1` The pH will be decided by strong acid, since dissociation of `CH_(3)COOH` is insignificant in presence of HCI due of common ion effecct. Thus pH solution of `0.1 MHCI=1` |
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