Equilibrium Question 749

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 $



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