wiz-icon
MyQuestionIcon
MyQuestionIcon
3
You visited us 3 times! Enjoying our articles? Unlock Full Access!
Question

Calculate the amount of (NH4)2SO2 in grams which must be added to 500 ml of 0.2 M NH4OH to yield a solution of pH=9. Kb(NH3)=2×105

A
3.24 g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4.24 g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
13.2 g
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
6.24 g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 13.2 g
Given, pH=9, Kb(NH3)=2×105, [NH3]=0.02 M, V=500 ml
let 'a' millimoles of NH+4 is added to a solution having millimoles
NH4OH=500×0.2=100[NH+4]=[salt]=a500 and [NH4OH]=[Base]=100500
pH+pOH=14pOH=149=5
Using Henderson equation,
pOH=logKb+log[NH+4][NH4OH] 5=log2×105+loga500100500
5=5log2+loga500100500
log2=loga500100500
a=200 millimoles = 0.2 mol
(NH4)2SO4 added=a2=0.1 mol W(NH4)2SO4=number of mole×molar mass of (NH4)2SO4
=0.1×132=13.2 g

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