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

For 0.50 M aqueous solution of sodium cyanide, (pKb of CN is 4.70), calculate
(i) Hydrolysis constant
(ii) Degree of hydrolysis
iii) pH
log( 3.15 )=0.498
Antilog (-9.3)=5.01×1010

A
2×105,6.3×103,11.5
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2×103,6.3×103,11.5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
4×103,6.3×103,9.5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2×103,4.2×103,11.5
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2×105,6.3×103,11.5
Salt hydrolysis of NaCN is
NaCN+H2OHCN+Na++OH
Given,
pKb=4.70

For conjugate acid-base pair,
pKa+pKb=14pKa=144.70pKa=9.3pKa=log Ka9.3=log KaKa=5.01×1010Kh=KwKaKh=1×10145.01×1010Kh=2×105

(ii) Degree of hydrolysis,
h=(Kh/c)=(2×105/0.5)

=6.3×103

(iii) [OH]=ch=0.5×6.3×103

[OH]=3.15×103
pOH=log [OH]
pOH=log(3.15×103)
pOH=3log(3.15)
pOH=30.498=2.501

pH+pOH=14
pH=142.5=11.5

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon