If to a 100 mL of solution of Q. 100, 0.002mol of HCl is added, its pH changes to - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
1.5 M
B
2.0
C
3.0
D
4.0

Solution:

In the solution of $Q.100, [{CH}_{3} COOH] = [{CH}_{3} COONa] = 0.01 M$

$n ({CH}_{3} COOH) = n ({CH}_{3} COONa) = (0.01 M) (0.1 {dm}^{3}) = 0.001 mol$

On adding 0.002l, $mol HCl$ ,the entire CH3COONa is replaced by CH3COOH. Thus, solution contains 0.002 $mol$ acetic acid and 0.001 $mol HCl$ . The solution concentrations are

$[{CH}_{3} COOH] = 0.02 M and [HCl] = 0.01 M$

The major source of $H^{+}$ in the solution will be $HCl$ .

Hence, $pH = - \log (0.01) = 2$

If to a 100 mL of solution of Q. 100, 0.002 $mol$ of $HCl$ is added, its pH changes to

Solution:

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