Equilibrium Question 693
Question: Calculate the amount of $ {{(NH_4)} _2}SO_4 $ in grams which must be added to 500 ml of 0.2 M $ NH_3 $ , to yield a solution of pH=9, $ K _{b} $ for $ NH_3=2\times {10^{-5}} $
Options:
A) 3.248 g
B) 4.248 g
C) 1.320 g
D) 6.248 g
Show Answer
Answer:
Correct Answer: C
Solution:
$ pOH=-\log K_{b}+\log \frac{[NH_4^{+}]}{[NH_4OH]} $ Let ‘a’ millimoles of $ N{H^{+}}_4 $ is added to a solution having millimoles of $ NH_4OH=500\times 0.2=100 $
$ \therefore [NH_4^{+}]=[salt]=\frac{a}{500} $ and $ [NH_4OH] $
$ =[Base]=\frac{100}{500} $ Given $ K_{b} $ for $ NH_4OH=2{{\times }^{-5}} $ and $ pH=9 $
$ \therefore 5=-\log 2\times {10^{-5}}+\log \frac{a/500}{100/500} $
$ \therefore a=200 $ millimoles = 0.2 mol Moles of $ {{(NH_4)}_2}SO_4 $ added $ =\frac{a}{2}=0.1 $ mol
$ \therefore W{{(NH_4)}_2}SO_4=0.1\times 132=1.32 $
BETA
