What will be the $p H$ of an aqueous solution of $0.1 M$ ammonium formate $25^{o} C$ ?
$Given : p K_{a} (H C O O H) = 3.8 a n d p K_{b} (N H_{4} O H) = 4.8$

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6.5

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4.5

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5.5

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Solution

The correct option is B 6.5
Ammonium formate is a salt of weak acid formic acid and weak base ammonium hydroxide.
here,
$C / K_{h} < 100$ so we can take $1 - h \approx 1$
So, the expression for the $p H$ of ammonium formate is: $p H = \frac{1}{2} (p K_{w} + p K_{a} - p K_{b})$ .
Substitute values in the above expression:
$p H = \frac{1}{2} (14 + 3.8 - 4.8) p H = 6.5$
Option C is correct.

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Similar questions

Q. What will be the

p H

of an aqueous solution of

0.1 M

ammonium formate

25^{o} C

Given : p K_{a} (H C O O H) = 3.8 a n d p K_{b} (N H_{4} O H) = 4.8

Q. Calculate the pH of an aqueous solution of

1 M

ammonium formate, assuming complete dissociation.

(p K_{a} = 3.8, p K_{b} = 4.8)

Calculate the pH of an aqueous solution of $1.0 M$ ammonium formate assuming complete dissociation. ( $p_{K a}$ of formic acid = 3.8 and $p_{K b}$ of ammonia = 4.8)

Q. Find the pH of an aqueous solution of 1.0 M ammonium formate assuming complete dissociation.

({pK}_{a} {of formic acid = 3.8 and pK}_{b} of ammonium hydroxide = 4.8.)

Q. Calculate pH of an aqueous solution of 1M

H C O O N H_{4}

assuming complete dissociation.

p K_{a} (H C O O H) = 3.8 a n d p K_{b} = 4.8

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Salt of Weak Acid and Weak Base

Standard XI Chemistry

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What will be the pH of an aqueous solution of 0.1 M ammonium formate 25oC? Given : pKa(HCOOH)=3.8 and pKb(NH4OH)=4.8

What will be the $p H$ of an aqueous solution of $0.1 M$ ammonium formate $25^{o} C$ ?
$Given : p K_{a} (H C O O H) = 3.8 a n d p K_{b} (N H_{4} O H) = 4.8$