Equilibrium Question 402
Question: By adding $ 20,ml $ $ 0.1,N,HCl $ to $ 20,ml $ $ 0.001,N, $ $ KOH, $ the pH of the obtained solution will be [KCET 2000]
Options:
A) 2
B) 1.3
C) 0
D) 7
Show Answer
Answer:
Correct Answer: B
Solution:
20 ml. of 0.1 NHCl = $ \frac{0.1}{1000}\times 20g $ eq. = $ 2\times {10^{-3}}g $ eq. 20ml. of 0.001 KOH = $ \frac{0.001}{1000}\times 20,gm $ eq. = 2 × $ {10^{-5}}g $ eq.
$ \therefore $ HCl left unneutralised = $ 2({10^{-3}}-{10^{-5}}) $
$ =2\times {10^{-3}}(1-0.01) $ = $ 2\times 0.99\times {10^{-3}} $
$ =1.98\times {10^{-3}}g,eq. $ Volume of solution = 40 ml. \ [HCl] = $ \frac{1.98\times {10^{-3}}}{40}\times 1000M $ = $ 4.95\times {10^{-2}} $
$ \therefore $ pH = $ 2- $ log 4.95 = 2 - 0.7 = 1.3.
