Explore topic-wise InterviewSolutions in .

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.

151.

Equal volumes of three acid solutions of `pH 3, 4` and `5` are mixed in a vessel. What will be the `H^(+)` ion concentration in the mixture?A. `3.7xx10^(-4) M`B. `3.7xx10^(-3) M`C. `1.11xx10^(-3) M`D. `1.11xx10^(-4) M`

Answer» Correct Answer - A
The problem is valid only if strong acids are mixed. After mixing equal volumes of three acids, total volume `=3V`.
`:. [H^(+)]` after mixing
`=(10^(-3)xxV)/(3V)+(10^(-4)xxV)/(3V)+(10^(-5)xxV)/(3V)=(1.11xx10^(-3))/(3)`
`=3.7xx10^(-4)`
Discussion
152.

Equal volumes of three acid solutions of `pH 3, 4` and `5` are mixed in a vessel. What will be the `H^(+)` ion concentration in the mixture?A. `3.7xx10^(-3) M`B. `1.11xx10^(-3) M`C. `1.11xx10^(-4) M`D. `3.7xx10^(-4) M`

Answer» Correct Answer - D
`[H^(+)]=10^(-3) M, [H^(+)]=10^(-4) M, [H^(+)]= 10^(-5) M` for the given acids.
`M_(mix)V_(mix)=M_(1)V_(1)+M_(2)V_(2)+M_(3)V_(3)`
` M_(mix)xx3=10^(-3)xx1+10^(-4)xx1+10^(-5)xx1`
`M_(mix)=(10^(-5)[100+10+1])/(3)=(111xx10^(-5))/(3)`
`=37xx10^(-5) M=3.7xx10^(-4) M`
Discussion
153.

`HA` is a weak acid and `BOH` is a weak base. For which of the following salts the extent of hydrolysis is independent of the concentration of the salt in its aqueous solutionA. `NaA`B. `NaB`C. `BCl`D. `BA`

Answer» Correct Answer - D
The extent of the hydrolysis `(h)` of the salt made by weak acid and weak base is independent of the concentration of salt.
`underset(weak acid)(HA)+underset(weak base)(BOH) rarr BA+H_(2)O`
`K_(h)=` hydrolysis constant `=(K_(w))/(K_(a)K_(b))`
Discussion
154.

A weak acid `(HA)` after treatment with `12 mL` of `0.1 M` strong base `(BOH)` has a `pH` of `5`. At the end point , the volume of same base required is `27 mL`. `K_(a)` of acid is `(log 2=0.3)`A. `1.8xx10^(-5)`B. `8xx10^(-6)`C. `1.8xx10^(-6)`D. `8xx10^(-5)`

Answer» Correct Answer - B
mmoles of `HA` taken `=27xx0.1=2.7`
`underset(1.5)underset(2.7)(HA)+underset(-)underset(1.2)(OH^(-) )rarr underset(1.2)underset()(A^(-))+H_(2)O`
`pH =pK_(a)+log``([A^(-)])/([HA])implies 5=pK_(a)+log``((1.2)/(1.5))`
`=pK_(a)+log``(4)/(5)`
`:. pK_(a)=5.1 implies K_(a)=80xx10^(-6)`
Discussion
155.

The concentration of `H^(+)` ion in a `0.2 M` solution of `HCOOH` is `6.4xx10^(-3) "mole" L^(-1)`. To this solution `HCOONa` is added so as to adjust the concentration of `HCOONa` to one mole per litre. What will be the `pH` of this solution? `K_(a)` for `HCOOH` is `2.4xx10^(-4)` and the degree of dissociation of `HCOONa` is `0.75`A. `3.19`B. `4.19`C. `5.19`D. `6.19`

Answer» Correct Answer - B
Assuming that the addition of `HCOONa` suppresses the ionization of `HCOOH`, we can use the expression
`pH=pK_(a)+log``(["Salt"])/([Acid])`
to compute `pH` of the solution, since salt is `75%` dissociated we will get,
`pH= -log(2.4xx10^(-4))+log``(0.75)/(0.2)`
`=3.62+0.57=4.19`
Discussion
156.

If `50ml` of `0.2 M KOH` is added to `40 ml` of `0.05 M HCOOH`, the `pH` of the resulting solution is (`K_(a)=1.8xx10^(-4)`)A. `3.4`B. `7.5`C. `5.6`D. `3.75`

Answer» Correct Answer - A
`pH=-logK_(a)+log``(["Salt"])/([Acid])`
`["Salt"]=(0.2xx50)/(1000)=0.01`, `[Acid]=(0.5xx40)/(1000)=0.02`
`pH=-log(1.8xx10^(-4))+log``(0.01)/(0.02)`
`pH=4-log(1.8)+log 0.5`
`pH=4-log (1.8)-0.301`
`pH=3.4`
Discussion
157.

`pH` of a solution of `10 ml`. `1 N` sodium acetate and `50 ml 2N` acetic acid `(K_(a)=1.8xx10^(-5))` is approximatelyA. `4`B. `5`C. `6`D. `7`

Answer» Correct Answer - A
`pH=pK_(a)+log``([Sal t])/([Acid])`
`pH= -log(1.8xx10^(-5))+log``([10])/([100])`
`=-log 1.8+5+log 10^(-1)`
`= -0.2553+5-1=3.7447` or `=4`
Discussion
158.

Which of the following have maximum `pH`?A. Black coffeeB. bloodC. Gastric juiceD. Saliva

Answer» Correct Answer - B
Black coffee `rarr 5.0`
Blood `rarr 7.4`
Gastric juice `rarr 1.8-2.0`
Saliva `rarr 6.8`
Discussion
159.

`10^(-6)M HCl` is diluted to `100` times. Its `pH` is:A. `6.0`B. `8.0`C. `6.95`D. `9.5`

Answer» Correct Answer - C
New concentration of `HCl=(10^(-5))/(100)=10^(-8) M`
`[H^(+)]=10^(-7)+10^(-8)` (approximately)
(Little less than `10^(-7)` from water).
Discussion
160.

A solution has `pH=5`, it is diluted `100` times, then it will becomeA. neutralB. basicC. unaffectedD. more acidic

Answer» Correct Answer - A
`pH=5` means `[H^(+)]=10^(-5)M`.
After dilution `[H^(+)]=10^(-5)//100=10^(-7)M`
`[H^(+)]` from `H_(2)O` cannot be neglected.
Total `[H^(+)]=10^(-7)+10^(-7)=2xx10^(-7)`
`pH=7-0.3010=6.6990=7` (neutral)
Discussion
161.

An acid solution of `pH=6` is diluted `1000` times, the `pH` of the final solution isA. `6.99`B. `6.0`C. `3.0`D. `9.0`

Answer» Correct Answer - A
`pH= 6 means [H^(+)]=10^(-6) M`. After dilution, the hydrogen ion concentration becomes `10^(-9) M`. Under such conditions, the hydrogen ions obtained from water cannot be neglected.
` :.` Total `[H^(+)]=10^(-9)+10^(-7)` (approx).
` =10^(-7)(10^(-2)+1)=10^(-7) (1.01)`
[The contribution of `H^(+)` from water will not be exactly `10^(-7)` but still we can make an approximation in an objective problem and take it as `10^(-7)`]
`pH= -log[H^(+)]= -1.01xx10^(-7)=7-0.0043=6.9957`
Discussion
162.

`10^(-6) M NaOH` is diluted by `100` times. The `pH` of diluted base isA. Between `6` and `7`B. Between `10` and `11`C. Between `7` and `8`D. Between `5` and `6`

Answer» Correct Answer - C
After dilution, `[OH^(-)]=10^(-8)M`
or `[H^(+)]=10^(-6) M. pH` cannot be `6` because total
`[OH^(-)]=10^(-7)+10^(-8)=10^(-7)(1+.01)`
`=10^(-7)xx1.1` or `[H^(+)]=10^(-14)//(10^(-7)xx1.1)`
`9.09xx10^(-8)`
`:. pH=9-log 9.09=8-0.9546=7.0454`
Discussion
163.

When rain is accompained by a thunderstorm, the collected rain water will have a `pH`:A. Influenced by occurrence of thunder stormB. Depends upon the amount of dust in waterC. Slightly lower than that of rain water without thunderstormD. Slightly higher than that when thunderstorm is not there

Answer» Correct Answer - C
Thunderstorm produces acidic oxides which on dissolution in water form acidic rains, i.e., `pHlt7`.
Discussion
164.

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

Answer» Correct Answer - A
`pH` of `1xx108 M HCl` is `-7`(`HCl` is acid)
Discussion
165.

The species among the following which can act as an acid and as a base isA. `HSO_(4)^(-)`B. `SO_(4)^(2-)`C. `H_(3)O^(+)`D. `Cl^(-)`

Answer» Correct Answer - A
The species which can accept as well as donate `H^(+)` can act both as an acid and as a base.
`underset(base)(HSO_(4)^(-))+H^(+)hArrH_(2)SO_(4)`
`underset(acid)(HSO_(4)^(-))+H^(+)hArrS_(4)^(2-)+H^(+)`
Discussion
166.

Solubility of `BaF_(2)` in a solution of `Ba(NO_(3))_(2)`, will be represented by the concentration term:A. `[Ba^(2+)]`B. `[F^(-)]`C. `1//2[F^(-)]`D. `2[NO_(3)^(+)]`

Answer» Correct Answer - C
Let `S` is the solubility of `BaF_(2)` in a solution of `BaNO_(3)`
Then `K_(SP)=[Ba^(2+)][F^(-)]^(2)`
Then `[F^(-)]=2S`,
Then `(1)/(2)[F^(-)]=S`
Discussion
167.

The `pH` of `Ba(OH)_(2)` solution is `13`. The number millimoles of `Ba(OH)_(2)` present in `10 ml` of solution would beA. `1.00`B. `0.50`C. `10.00`D. `15.00`

Answer» Correct Answer - B
`[OH^(-)]=0.1` and conc. of `Ba(OH)_(2)=(0.1)/(2)`
`:.` The no. of milimoles of `Ba(OH)_(2)` present in `10 ml` soultion `=(0.1)/(2)xx10xx10^(-3)xx10^(3)=0.5`
Discussion
168.

`100 mL` of a buffer solution contains `0.1 M` each of weak acid `HA` and salt `NaA`. How many gram of `NaOH` should be added to the buffer so that it `pH` will be `6` ? (`K_(a)` of `HA=10^(-5)`).A. `4.19`B. `0458`C. `0.328`D. None

Answer» Correct Answer - C
For acidic buffer, `pH=pK_(a)+log``(0.1)/(0.1)`
`pH=pK_(a)= -log(10^(-5))=5`.
Rule: `ABA` (In acidic buffer (`A`), on addition of `SB(B)`, the concentration of `WA(A)` decreases and that of salt increases).
Let `x M` of `NaOH` is added.
`pH_(new)=5+log``((0.1+x)/(0.1-x))`
`6-5= log``((0.1+x)/(0.1-x))`
`((0.1+x)/(0.1-x)) = anti log (1)=10`
Solve for `x`:
`x=0.082 M=(0.082)/(1000)xx100`
`=0.0082 mol (100 mL)^(-1)`
`=0.0082xx40 g (100mL)^(-1)`
`0.328 g`
Discussion