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1851.

The solubility product constant of a metal carbonate `MCO_(3)` is `2 xx 10^(-12)` at `25^(@)C`. A solution is `0.1 M` in `M(NO_(3))_(2)` and it is saturated with `0.01 M CO_(3)`. Also the ionization constant of `CO_(2)` are : `K_(a_(1)) = 4 xx 10^(-7)` and `K_(a_(2)) = 5 xx 10^(-11)` at `25^(@)C`. The minimum pH that must be maintained to start any precipitation is

Answer» Correct Answer - 4
Discussion
1852.

`100 mL` of `0.3 N`-Acetic acid solution is mixed with same volume of `0.2 N` sodium hydroxide solution. Ionisation constant of acetic acid is `2 xx 10^(-5)` the pH of the mixture isA. `10`B. `7`C. `5`D. `9`

Answer» Correct Answer - C
`CH_(3)COOH + NaOH rarr CH_(3)COONa + H_(2)O`
`100 xx 0.3 , 100 xx 0.2`
`(CH_(3)COOH) = (100 xx 0.3) - (100 xx 0.2)`
`[CH_(3)COONa] = 100 xx 0.2`
`pKa = - log k_(a)`
Discussion
1853.

Assertion (A): `H_(2)SO_(4)` acts as a base in the presence of `HCIO_(4)`. Reason (R) : Perchloride acid is stronger acid than `H_(2)SO_(4)`.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.

Answer» Correct Answer - A
Both `(A)` and `(R)` are correct and `(R)` is the correct explanantion of `(A)`.
`{:(H_(2)SO_(4)+,HCIO_(4)rarr,CIO_(4)^(Theta)+,H_(3)SO_(4)^(o+),),(Base,"Acid",C_(B),C_(A),):}`
Discussion
1854.

How many gram of `CaC_(2)O_(4)` will dissolve in distilled water to make one litre of saturated solution ? `(K_(sp) =2.5xx10^(-9)` and its molecular mass is 128 )A. 0.0064 gB. 0.0128 gC. 0.0032 gD. 0.0640 g

Answer» Correct Answer - A
`CaC_(2)O_(4) hArr Ca^(2+)+C_(2)O_(4)^(2-)`
`K_(sp)=[Ca^(+2)][C_(2)O_(4)^(-2)]`
`K_(sp)=s^(2)`
`s=sqrt(K_(sp))=sqrt(2.5 xx 10^(-9))`
`=5 xx 10^(-5) mol L^(-1)`
Solubility of `g L^(-1) =5 xx 10^(-5) xx ` mol. wt.
`=5 xx 10^(-5) xx 128 =0.00640 g L^(-1)`
Discussion
1855.

`20 mL` of `0.2M` sodium hydroxide is added to `50 mL` of `0.2 M `acetic acid to give `70 mL` of the solution. What is the `pH` of this solution. Calculate the additional volume of `0.2M NaOh` required to make the `pH` of the solution `4.74`. (Ionisation constant of `CH_(3)COOh` is `1.8 xx 10^(-5))`

Answer» Correct Answer - D
`20 mL` of `0.2M` sodium hydroxide is added to `50 mL` of `0.2M` acetic acid to acid to given `70 mL` of the solution. An acidic buffer is formed.
`{:(NaOH+,CH_(3)COOHrarr,CH_(3)COONa+,H_(2)O,),(("millimoles"),,,,),(20xx0.2,50xx0.2,,,),(=4,=10,0,0,),(0,10-4=6,4,4,):}`
`["Acid"] = (7)/(70) ["Salt"] = (4)/(70)`
`pH = pK_(a) +"log" (["Salt"])/(["Acid"])`
`=- log (1.8 xx 10^(-5)) +log"(4)/(70) xx (70)/(6)`
`= 4.74 + "log"(2)/(3) = 4.56`
Rule ABA: When `NaOH` is added, the amount of acid decreases because hydroxide ions react with hydrogen ions to form undissociated water. Therefore, the amount of acetate increases.
Initial millimoles of acid `= 6`
New millimoles of acid `= 6 - (0.2V)`
New `["Acid"] = (6-(0.2V))/(70)`
Initial millimoles of salt `= 4`
New millimoles of salt `= 4 + (0.2 V)`
New `["Salt"] = (4+(0.2V))/(70)`
`pH = pK_(a) "log"(["Salt"])/(["Acid"])`
or `4.74 =- log (1.8 xx 10^(-5))`
`+log {(4+(0.2V))/(70)xx(70)/(6-(0.2V))}`
`- log (1.8 xx 10^(-5)) ~~ 4.74`
`:. 4.74 = 4.74 + "log" (4+(0.2V))/(6-(0.2V))`
or `0 = "log" (4+(0.2V))/(6-(0.2V))`
or `1 = (4+(0.2V))/(6-(0.2V))`
or `4 +(0.2V) = 6 - (0.2V)`
or `0.4 V = 2`
or `V =5mL`
Discussion
1856.

Assuming `H_(2)SO_(4)` to be completely ionised the `pH` of a `0.05M` aqueous of sulphuric acid is approximatelyA. `0.01`B. `0.005`C. `2`D. `1`

Answer» Correct Answer - D
`0.05 M_(H_(2)SO_(4)) = 2xx0.05 N_(H_(2)SO_(4)) = 0.1N`
`:. [H_(3)O^(o+)] = [0.1] = 10^(-1)`
`-log [H_(3)O^(o+)] =- log [10^(-1)] = 1`
`pH = 1`
Discussion
1857.

In water, the acid `HCIO_(4), HCI, H_(2)SO_(4)` and `HNO_(3)` exhibit the same strength as they are completely ionised in water (a base). This is called …… of the solvent water.A. StrengthB. CapacityC. Buffer effectD. Levelling effect

Answer» Correct Answer - D
Discussion
1858.

When pure water is saturated with `CaCO_(3)` and `CaC_(2)O_(4)` the concentration of calcium ion in the solution under equilibrium is `8xx10^(-5)` M. If the ratio of the solubility product of `CaCO_(3)` to that of `CaC_(2)O_(4)` is 3, what is the solubility of `CaCO_(3)` in pure water?A. `4.80xx10^(-8)`B. `9.60xx10^(-9)`C. `9.62xx10^(-8)`D. `6.93xx10^(-5)`

Answer» Correct Answer - d
Discussion
1859.

How many grams of `CaC_(2)O_(4)` (molecular weight = 128) on dissolving water will give a saturated solution `[K_(sp)(CaC_(2)O_(4)) = 2.5 xx 10^(-9) mol^(2) I^(-2)]`A. `0.0064 g`B. `0.1280g`C. `0.0128 g`D. `1.2800 g`

Answer» Correct Answer - A
Solubility of `CaC_(2)O_(4) = sqrt(K_(sp)) = sqrt(2.5 xx 10^(-9))`
`= 5 xx 10^(-5) mol L^(-1)`
`= 5 xx 10^(-5) xx 128 = 640 xx 10^(-5) = 0.0064 g`.
Discussion
1860.

Assuming `H_(2)SO_(4)` to be completely ionised the `pH` of a `0.05M` aqueous of sulphuric acid is approximatelyA. 0.01B. 0.005C. 2D. 1

Answer» Correct Answer - D
`0.05 M H_(2)SO_(4)` means that `[H^(+)]` concentration is `2xx0.05 M=0.1 M`
Discussion
1861.

During the neutralisation of an acid by a base, the end point refers to the completion of reaction. The detection of end point in acid-base neutralisation is usually made by an acid-base indicator. An acid-base indicator is itself a weak acid (Phenolphthalein) or a weak base (Methyl orange). At `50%` ionisation which depends on the medium. the anion furnished by an indicator (acid) cation furnished colour to solution at end point. For example, phenolphthalein dissociation `underset("Colourless")(HIn)toH^(+)+underset("Pink")(In^(-)),K_(HIn)=([H^(+)][In^(-)])/([HIn])` is favoured in presence of alkali and pink colour of phenolphthalein ion is noticed as soon as the medium changes to alkali nature. The end point of acid-base neutralisation not necessarily coincides with eqivalence point but it is closer to equivalence point. Also at equivalence point of acid-base neutralisation, pH is not necessarily equal to 7. Which among the followig statements are correct? (P) At equivalence point of NaOH AND HCI,pH=7 (Q) At equivalence point of NaOH and `CH_(3)COOH,pH gt 7` (R) At equivalence point of `NH_(4)OH` and `HCI,pH lt 7` (S) An indicator shows best results if equivalence point is within the pH range `pK_(In)pm1` (T) ) At equivalence point of `NH_(4)OH` and formic acid `pH lt 7`A. P,Q,R,S and TB. P,R,S and TC. P,S and TD. P,Q,R and T

Answer» Correct Answer - a
Discussion
1862.

The pH of `(N)/(100)` HCl would be approximatelyA. 1B. 1.5C. 2D. 2.5

Answer» Correct Answer - C
`0.01 M HCl = 10^(-2) M[H^(+)] , pH = 2`.
Discussion
1863.

In an experiment to determine the enthalpy of neutralisation of sodium hydroxide with sulphuric acid, `50 cm^(3)` of `0.4M` sodium hydroxide were titreated thermometrically with `0.25M` sulphuric acid. Which of the following plots gives the correct representation ?A. B. C. D.

Answer» Correct Answer - B
Meq.of `H_(2)SO_(4)` needed for 20 Meq.of NaOH=20. Thus volume of `H_(2)SO_(4)` needed `=V_(mL)` or `Vxx0.25xx2=20, :. V= 40mL`
Also temperature increases during neutralisation and becomes constant after neutralisation.
Discussion
1864.

In one litre of water `10^(-10)` moles of HCl were added. The pH of the solution will be approximately .A. 7B. 14C. 10D. 4

Answer» Correct Answer - A
Molar conc. Of HCl`=10^(-10)M`
`[H_(3)O^(+)]` ion already present in water `=10^(-7)M`
Net `[H_(3)O^(+)]=(10^(-7)+10^(-10))`
`=10^(-7) (1+10^(-3))=1.001 xx 10^(-7)`
`pH =-log (1.001 xx 10^(-7))= 7- log 1.001 ~~7`
Discussion
1865.

To an aqueous solution of `Ag_(2)CrO_(4)` above its own precipitate , `CrO_(4)^(2-)` ions are added in the form of solution. This results inA. increase in concentration of `Ag^(+)` ionsB. decrease in concentration of `Ag^(+)` ionsC. increase in the value of solubility productD. decrease in the value of solubility product.

Answer» Correct Answer - B
Equilibrium in solution of sparingly soluble `Ag_(2)CrO_(4)`
`Ag_(2)CrO_(4) hArr 2 Ag^(+) +CrO_(4)^(2-)`
Additon of `CrO_(4)^(2-)` ions will push the equilibrium in backward direction.
Discussion
1866.

A well is dry in a bed of rock containing fluorspar `(CaF_(2))`. If the well contains `20,000 L` of `H_(2)O`, what is the amount of `F^(-)` in it `K_(SP)` of `CaF_(2) = 4 xx 10^(-11)`A. 4.3 molB. 6.8 molC. 8.6 molD. 13.6 mol

Answer» Correct Answer - c
Discussion
1867.

Which of the following solutions have different `pH`? `100mL" of "0.2m" "HCl +100mL" of "0.4 M" "NH_(3)` `50mL" of "0.1 M" HCl +"50mL" of "03.2 M NH(3)` `10 mL of 0.3" M "HCl +100mL of 0.6 M NH_(3)`A. I &iiB. ii & iiiC. I & iiiD. all will have same `pH`.

Answer» (a) `{:(NH_(4)OH,+,HCl,rarr,NH_(4)Cl,+,H_(2)O),(0.2M,,0.1,,0.1,,0),(0.1,,0,,0.1,,0.1):}`
`pOH=pK_(b)`
(b) `{:(NH_(4)OH,+,HCl,rarr,NH_(4)Cl,+,H_(2)O),(0.1M,,0.05,,0,,0),(0.05,,0,,0.05,,):}`
`pOH=pK_(b)+log(((0.05)/(0.05)))=pK_(b)`
c `{:(NH_(4)OH,+,HCl,rarr,NH_(4)Cl,+,H_(2)O),(0.3,,0.15,,,,),(0.15,,0,,0.15,,):}`
`pOH=pK_(b)` so all solution have same `pH`.
Discussion
1868.

`2.5mL` of `(2)/(5)` weak monoacidic base (`K_(b)= 1xx10^(-12)` at `25^(@)C`) is tittrated with `(2)/(15)MHCI` in water at `25^(@)C`. The concentration of `H^(+)` at equivalence point is: `(K_(w)= 1xx10^(-14)at 25^(@)C)`A. `3.7xx10^(-14) M`B. `3.2xx10^(-7) M`C. `3.2xx10^(-2) M`D. `2.7xx10^(-2) M`

Answer» Correct Answer - d
Discussion
1869.

Upon titrating a solution of weak monoprotic acid with a weak monoacidic base solution at equivalent point:A. `pH` in general weak would increase if both solution are first diluted to `10` times the original volume & then titration is carries outB. `pH` in general would decrease if both solution are first diluted to `10` times the original volume & then titration is carried out.C. `pH` in general would remain same `(=7)` if both solutions are first diluted to `10` times the original volumes & then titration is carried outD. `pH` in general coule be less than , greater then or equal to `7`if both solutions are first diluted to`10` times the original volume & then titration is carried out.

Answer» Salt of `WA-WB` formed at equivalence point has `pH` independent of concentration and depends on `pK_(a)` and `pK_(b)` of `WA` and `WB` respectively.
Discussion