The dissociation constant of an acid $HA$ is $1 \times 10^{-5}$. The $pH$ of $0.1 \ M$ solution of the acid will be

Consider the dissociation equilibrium of the following weak acid $HA \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$. If the $pK_{a}$ of the acid is $4$,then the $pH$ of $10 \ mM$ $HA$ solution is . . . . . . . (Nearest integer)
[Given : The degree of dissociation can be neglected with respect to unity]

For a $0.1 \ M$ solution of a weak acid $HA$ $(K_a = 1.4 \times 10^{-5})$ in $2 \ L$ of solution,calculate the percentage of dissociation and the $pH$ of the solution.

What will be the $pH$ of a $0.1 \ M$ $NH_3$ solution?

In $20 \, mL$ of $0.4 \, M$ $HA$ solution,$80 \, mL$ of water is added. Assuming volume to be additive,the $pH$ of the final solution is ($K_a$ of $HA = 4 \times 10^{-7}, \log 2 = 0.3$).

If two acids of equimolar concentration are taken,then which option is correct?