Correct option is (b) (A) > (D) > (C) > (B)
Formulae used - \(\mathrm{pH}=-\log \mathrm{H}^{+}\)
\(\mathrm{pOH}=-\log \mathrm{OH}^{-}\)
\(\mathrm{pH}+\mathrm{pOH}=14\)
Calculation of pH of 0.01 M HCl -
0.01 M HCl contains \(10^{-2} \mathrm{M} \mathrm{H}^{+}\)
\(\mathrm{pH}=-\log 10^{-2}\)
\(\mathrm{pH}=2\)
\(\mathrm{pOH}=14-2\)
\(\mathrm{pOH}=12\)
Calculation of pH of 0.01 M NaOH -
\(0.01 \mathrm{M}\, \mathrm{NaOH}^{-2}\) contains \(10^{-2} \mathrm{M} \mathrm{OH}^{-}\)
\(\mathrm{pOH}=-\log 10^{-2}\)
\(\mathrm{pOH}=2\)
Calculation of pH of \(0.01 \mathrm{M}\, \mathrm{CH}_{3} \mathrm{COONa} \) -
\(\mathrm{CH}_{3} \mathrm{COONa}\) is a salt of weak acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) and strong base \((\mathrm{NaOH})\).
So, its pH will be greater than 7 and pOH will be less than 7 .
\(\mathrm{pOH}<7\) (but not less than 2 as it is not a strong acid but a salt of a strong acid).
Calculation of pH of 0.01 M NaCl -
NaCl is a salt of strong base \((\mathrm{NaOH})\) and strong acid \((\mathrm{HCl})\).
So, \(\mathrm{pH}=7\)
\(\mathrm{pOH}=7\)
So, the order of the following solutions in the decreasing order of \(\mathrm{pOH}=(\mathrm{A})> (D) > (C) > (B)\).
