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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.5

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

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

Join BYJU'S Learning Program

Explore more

Salt of Weak Acid and Weak Base

Standard XIII Chemistry

Join BYJU'S Learning Program

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 $25^{o} C$ ?
$Given : p K_{a} (H C O O H) = 3.8 a n d p K_{b} (N H_{4} O H) = 4.8$