Equilibrium Question 314
Question: Calculate the amount of $ {{(NH_4)}_2}SO_4 $ in grams which must be added to 500 ml of $ 0.200,M,NH_3 $ to yield a solution with $ pH=9.35 $
$ (K_{b} $ for $ NH_3=1.78\times {10^{-5}}) $ [UPSEAT 2001]
Options:
A) 10.56 gm
B) 15 gm
C) 12.74 gm
D) 16.25 gm
Show Answer
Answer:
Correct Answer: A
Solution:
$ pOH=pKb+\log \frac{[Salt]}{[Base]} $
$ 14-9.35=-\log (1.78\times {10^{-5}})+\log \frac{[Salt]}{100} $
$ [Salt]=79.9 $
Þ $ \frac{w}{132}\times 1000=79.9 $
Þ $ w=10.56 $
