Question

Which mixture forms a buffer when dissolved in 1L of water?
A. 0.2 mol NaOH + 0.2 mol HBr
B. 0.2 mol NaCl + 0.3 mol HCl
C. 0.4 mol $HN{{O}_{2}}$ + 0.2 mol NaOH
D. 0.5 mol $N{{H}_{3}}$ + 0.5 mol HCl

Answer 1

To solve this question you should have basic knowledge about Buffer solutions. Buffer solutions are solutions which resist change in pH on dilution of small amounts of acid or alkali.

Complete answer:
The solutions which resist change in pH on dilution or with the addition of small amounts of acid or alkali are called Buffer solution. Buffer solutions of known pH can be prepared from the knowledge of $p{{K}_{a}}$ of the acid or $p{{K}_{b}}$ of base and by controlling the ratio of the salt and acid or salt and base. A mixture of acetic acid and sodium acetate acts as buffer solution around pH 4.75 and a mixture of ammonium chloride and ammonium hydroxide act as a buffer around pH 9.25.
A. 0.2 mol NaOH + 0.2 mol HBr will not form a buffer solution because it is a mixture of a strong base and a strong acid.
B. 0.2 mol NaCl + 0.3 mol HCl will not form a buffer solution because it is a mixture of a neutral salt and a strong acid.
C. 0.4 mol $HN{{O}_{2}}$ + 0.2 mol NaOH will form a buffer solution because it is a mixture of a weak acid and its conjugate base. 0.2 mol of NaOH will neutralize 0.2 mol of $HN{{O}_{2}}$. Hence, you will be left with 0.2 mol of $HN{{O}_{2}}$ and 0.2 mol of $NaN{{O}_{2}}$ (or, more specifically,$NO_{2}^{-}$)
The pH will be equal to the $p{{K}_{a}}$ of $HN{{O}_{2}}$.
Hence, 0.4 mol $HN{{O}_{2}}$ + 0.2 mol NaOH will form a buffer as it is a mixture of weak acid and its conjugate base.
D. 0.5 mol $N{{H}_{3}}$ + 0.5 mol HCl would have formed a buffer if you had added less HCl; however, you are adding so much HCl that it completely protonated $N{{H}_{3}}$ to $NH_{4}^{+}$ .Hence, you will be left with 0.5 mol of $NH_{4}^{+}$ and 0.5 mol of $C{{l}^{-}}$.

Hence, option C is the correct answer.

Note: You should have basic knowledge about strong acid, strong base and weak acid, weak base and also conjugates of acid and base.
You should remember $p{{K}_{a}}=-[\log {{K}_{a}}]$ and $p{{K}_{b}}=-[\log {{K}_{b}}]$
Conjugate pairs of acid and base differ only by one proton.