Calculate the molar ratio of a weak acid $HA$ $(K_a=10^{-6})$ and its salt with a strong base,so that the $pH$ of the buffer solution is $6$.

The dissociation constant of a weak acid is $1 \times 10^{-4}.$ In order to prepare a buffer solution with a $pH = 5,$ the $[Salt]/[Acid]$ ratio should be

$A$ buffer solution is a mixture of

Henderson's equation is $pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}$. If the acid is half-neutralized,the value of $pH$ will be: $[pK_a = 4.30]$

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

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