wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Calculate the pH of an aqueous solution of 1.0 M ammonium formate assuming complete dissociation. (pKa of formic acid = 3.8 and pKb of ammonia = 4.8)


A

7.2

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

7.5

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

6

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

6.5

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

The correct option is D

6.5


This is a bit of a strange situation isn't it?

NH+4 + HCOO + H2O NH4OH + HCOOH

So here we have a mixture of a weak acid and a weak base. So we'll have to use the corresponding hydrolysis equation. Before that, we make a note that pKa of formic acid = 3.8 and pKb of ammonia = 4.8 are given.

This makes it easier guess because if we separately write the hydrolysis equations for the weak acid and the weak base, we might figure out a way to get [H+], which is in a way, what we are looking for

Kh = [NH4OH][HCOOH][NH+4][HCOO] ............(1)

Ka = [HCOO][H][HCOOH] ............(2)

Kb = [OH][NH4][NH4OH] ............(3)

Kh = [OH][H]Kb.Ka = KwKaKb ............(4)

Moreover, in the very first Ammonium Formate hydrolysis reaction,0

[NH4OH] = [HCOOH] and [NH+4] = [HCOO]

Substituting the above in Eq. (1)

Kh = [HCOOH]2[HCOO]2

Using the above and combining (2) and (4), we have

KwKaKb = [H]2Ka2

Or [H+]2 = KaKwKb

Taking Log10 on both sides of the above equation, we have

2pH = pK+pKapKb = [14+3.84.8]

Or pH = 6.5


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon