Buffer Solutions

An acidic buffer is prepared by mixing a weak acid and its salt with a strong base (e.g. acetic acid and sodium acetate), which resists changes in pH on the acidic side.

50 mL of 0.1 M NaOH + 25 mL of 0.1 M $CH_3COOH$: 5 mmol NaOH and 2.5 mmol acid → 2.5 mmol NaOH left. Basic due to excess NaOH, not a buffer. 100 mL of 0.1 M $CH_3COOH$ + 100 mL of 0.1 M NaOH: 10 mmol each → only 10 mmol $CH_3COONa$ (salt hydrolysis), not a buffer. 100 mL of 0.1 M HCl + 200 mL of 0.1 M $NH_4OH$: 10 mmol HCl and 20 mmol $NH_4OH$ → 10 mmol $NH_4OH$ + 10 mmol $NH_4Cl$. This is a basic buffer. 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH: 10 mmol each → neutral NaCl solution.

An acidic buffer is a weak acid with its salt.
$HClO_4$ is a strong acid, so $HClO_4$ and $NaClO_4$ cannot act as an acidic buffer.

A buffer contains a weak acid and its conjugate base (or a weak base and its conjugate acid).
Added $H^+$ combines with the conjugate base to form the unionised weak acid, e.g. $CH_3COO^- + H^+ \rightarrow CH_3COOH$.
Added $OH^-$ reacts with the weak acid: $CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$.
Since the added acid or alkali is converted into unionised acid or base, the pH stays almost constant.

This is a basic buffer of a weak base ($NH_3$) and its salt ($NH_4^+$).
$pK_b = -\log(1.8\times10^{-5}) = 5 - \log1.8 = 5 - 0.26 = 4.74$
Henderson equation: $pOH = pK_b + \log\frac{[\text{salt}]}{[\text{base}]}$
$pOH = 4.74 + \log\frac{0.20}{0.30} = 4.74 + (\log2 - \log3) = 4.74 + (0.301 - 0.477)$
$pOH = 4.74 - 0.176 = 4.56$
$pH = 14 - pOH = 14 - 4.56 = 9.44$
This matches the value 9.43 among the choices.

$B^-$ is the conjugate base of the weak acid HB; $pK_b = -\log(10^{-10}) = 10$
Henderson equation for the base: $pOH = pK_b + \log\frac{[HB]}{[B^-]}$
Since $[B^-] = [HB]$, the log term is $\log1 = 0$, so $pOH = pK_b = 10$
$pH = 14 - pOH = 14 - 10$
pH = 4
Check: $K_a$ of HB = $\frac{K_w}{K_b} = \frac{10^{-14}}{10^{-10}} = 10^{-4}$, so $pH = pK_a = 4$.

The solution is an acidic buffer: $CH_3COOH \rightleftharpoons CH_3COO^- + H^+$
$CH_3COONa$ is fully dissociated, giving $[CH_3COO^-] \approx 0.20$ M; because of this common ion the ionisation of the acid is suppressed, so $[CH_3COOH] \approx 0.10$ M.
$K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}$
$[H^+] = \frac{K_a[CH_3COOH]}{[CH_3COO^-]} = \frac{1.8\times10^{-5}\times0.10}{0.20}$
$[H^+] = 9.0\times10^{-6}$ mol/L

A buffer solution resists a change in pH on the addition of a small amount of acid or base.
An acidic buffer is a mixture of a weak acid and its salt with a strong base; a basic buffer is a mixture of a weak base and its salt with a strong acid.
$HNO_2$ is a weak acid and $NaNO_2$ is its salt with the strong base NaOH, so this pair constitutes an acidic buffer.
Added $H^+$ is removed by $NO_2^-$ ($NO_2^- + H^+ \rightarrow HNO_2$), and added $OH^-$ is removed by $HNO_2$ ($HNO_2 + OH^- \rightarrow NO_2^- + H_2O$).
NaOH is a strong base, and $HNO_3$ and HCl are strong acids; a strong acid or strong base with its salt does not form a buffer.
So the buffer is $HNO_2$ and $NaNO_2$.

This is a basic buffer. Henderson equation: $pOH = pK_b + \log\dfrac{[salt]}{[base]}$
Here [salt] = [base] = 0.1, so $\log 1 = 0$ and $pOH = pK_b$.
$pOH = 14 - pH = 14 - 9.25 = 4.75$
So $pK_b = 4.75$.