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Standard XI

Chemistry

pH the Power of H

At 25∘C, the ...

Question

At $25^{\circ} C$ , the ionization constants of $C H_{3} C O O H$ and $N H_{3}$ are almost equal. The pH of a solution of $0.01 M C H_{3} C O O H$ is 4.0 at $25^{\circ} C$ . The pOH of 0.01 $M N H_{4} O H$ at the same temperature will be :

3.0

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4.0

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10.0

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10.5

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Solution

The correct option is C
10.0

Since $K_{a} = K_{b^{'}} [H_{3} O^{+}]$ in $0.01 M C H_{3} C O O H = [O H^{-}] i n 0.01 M N H_{4} O H$ .
Hence, pOH of $N H_{4} O H_{(a q)} = p H o f C H_{3} C O O H_{(a q)}$
$= 4; p H o f N H_{4} O H = 14 - 4 = 10$

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Similar questions

At $25^{\circ} C$ , the ionization constants of $C H_{3} C O O H$ and $N H_{3}$ are almost equal. The pH of a solution of $0.01 M C H_{3} C O O H$ is 4.0 at $25^{\circ} C$ . The pOH of 0.01 $M N H_{4} O H$ at the same temperature will be :

Q. Degree of dissociation of

C H_{3} C O O H

and

N H_{4} O H

are the same. If

0.01

M solution of

C H_{3} C O O H

has

p H = 4.0

; then

p H

0.01

N H_{4} O H

will be:

Q. At

25^{o} C

, the dissociation constants of

{C H}_{3} C O O H

and

{N H}_{4} O H

in aqueous solution are almost the same. The pH of a solution of

0.01 N

{C H}_{3} C O O H

4.0

25^{o} C

. The pH of

0.01 N

{N H}_{4} O H

solution at the same temperature would be:

Q. At

25^{0} C

dissociation constants of acid HA and base BOH in aqueous solution are same. the pH of 0.01 M solution of HA is 5. The pOH of

10^{- 4}

M solution of BOH at the same temperature is:

Q. The correct statements among the following:

a) At

25^{\circ} C

, pH = pOH = 7

b) pH of an acid solution always less than 7 at

25^{\circ} C

c) pH of 0.001 M

C H_{3} C O O H

solution is equal to 3 at

25^{\circ} C

d) pH of 0.05 M

H_{2} S O_{4}

is equal to pH of 0.1 M H Cl solution

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At 25∘C, the ionization constants of CH3COOH and NH3 are almost equal. The pH of a solution of 0.01 M CH3COOH is 4.0 at 25∘C. The pOH of 0.01 M NH4OH at the same temperature will be :

The correct option is C 10.0 Since Ka=Kb′[H3O+] in 0.01MCH3COOH=[OH−] in 0.01 MNH4OH. Hence, pOH of NH4OH(aq)=pH of CH3COOH(aq) =4;pH of NH4OH=14−4=10

At $25^{\circ} C$ , the ionization constants of $C H_{3} C O O H$ and $N H_{3}$ are almost equal. The pH of a solution of $0.01 M C H_{3} C O O H$ is 4.0 at $25^{\circ} C$ . The pOH of 0.01 $M N H_{4} O H$ at the same temperature will be :

The correct option is C
10.0

Since $K_{a} = K_{b^{'}} [H_{3} O^{+}]$ in $0.01 M C H_{3} C O O H = [O H^{-}] i n 0.01 M N H_{4} O H$ .
Hence, pOH of $N H_{4} O H_{(a q)} = p H o f C H_{3} C O O H_{(a q)}$
$= 4; p H o f N H_{4} O H = 14 - 4 = 10$