JEE Main · 2020 · 06 Sep 2020 · Shift 1 · Equilibrium

Since $\text{pH}+\text{pOH}$ is fixed, the most acidic solution has the lowest pH and so the highest pOH, and the ranking by pOH simply reverses the ranking by pH. Here $0.01\text{M}$ $\mathrm{HCl}$, a strong acid, is the most acidic and tops the pOH order; $0.01\text{M}$ $\mathrm{NaCl}$ is neutral; $0.01\text{M}$ $\mathrm{CH}{}_3\mathrm{COONa}$, the salt of a weak acid, is mildly basic; and $0.01\text{M}$ $\mathrm{NaOH}$, a strong base, is the least acidic with the lowest pOH. The decreasing order of pOH is therefore $\mathrm{HCl}>\mathrm{NaCl}>\mathrm{CH}{}_3\mathrm{COONa}>\mathrm{NaOH}$.