The $ pH$ of $ 0.1$ $M$ acetic acid is $3$, the dissociation constant of acid will be
$1.0 \times {10^{ - 4}}$
$1.0 \times {10^{ - 5}}$
$1.0 \times {10^{ - 3}}$
$1.0 \times {10^{ - 8}}$
If the $pKa$ of lactic acid is $5$,then the $pH$ of $0.005$ $M$ calcium lactate solution at $25^{\circ}\,C$ is $........\times 10^{-1}$ (Nearest integer)
$50\ ml$ of $0.02\ M$ $NaHSO_4$ is mixed with $50$ $ml$ of $0.02\ M\ Na_2SO_4$. Calculate $pH$ of the resulting solution.$[pKa_2 (H_2SO_4) = 2]$
The hydrogen ion concentration in weak acid of dissociation constant ${K_a}$ and concentration $c$ is nearly equal to
A $0.1\, M$ solution of $HF$ is $1\%$ ionized. What is the $K_a$
${K_b}$ of $N{H_4}OH = 1.8 \times {10^{ - 5}}$ calculate $pH$ of $0.15$ $mol$ $N{H_4}OH$ and $0.25$ $mol$ $N{H_4}OH$ containing solution.