Calculate $pH$ of the solution obtained by mixing $50 \ mL$ of $0.2 \ M$ $NH_4Cl$ solution and $75 \ mL$ of $0.1 \ M$ $NaOH$ solution. Given $pK_b$ for aqueous $NH_3$ is $4.74$.

$A$ buffer that is a mixture of acetic acid $(K_a = 2 \times 10^{-5})$ and potassium acetate has $pH = 5.18$. The $\frac{[CH_3COO^{-}]}{[CH_3COOH]}$ ratio in this buffer is approximately:

$A$ student needs to prepare a buffer solution of propanoic acid and its sodium salt with $pH$ $4$. The ratio of $\frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]}$ required to make the buffer is .....
Given: $K_{a}(CH_{3}CH_{2}COOH) = 1.3 \times 10^{-5}$

Calculate the $pH$ of a buffer solution containing $0.04 \ M$ $NaF$ and $0.02 \ M$ $HF$ $[pK_a = 3.142]$.

The $pH$ of a simple sodium acetate buffer is given by $pH = pK_a + \log \frac{[Salt]}{[Acid]}$. $K_a$ of acetic acid $= 1.8 \times 10^{-5}$. If $[Salt] = [Acid] = 0.1 \, M$,the $pH$ of the solution would be about

Assertion $(A)$: The $pH$ of a buffer solution containing equal moles of acetic acid and sodium acetate is $4.8$ ($pK_a$ of acetic acid is $4.8$).
Reason $(R)$: The ionic product of water at $25^{\circ} C$ is $10^{-14} \ mol^2 \ L^{-2}$. The correct answer is