MCQ
Given $pH $ of a solution $ A$ is $3$ and it is mixed with another solution $B$ having $pH\, 2$. If both mixed then resultant $pH$ of the solution will be
- A$3.2$
- ✓$1.9$
- C$3.4$
- D$3.5$
$[H+]_A = 10^{-3} M.$
$pH $ of the solution $B = 2$
$[H^+]_B =10^{-2} M$
$[H^+] = 10^{-3} + 10^{-2} = 10^{-3} + 10 × 10^{-3} = 11 \times 10^{-3}$.
$pH = -log(11 \times 10^{-3}) = 3 -log 11$ $= 3 -1.04 = 1.95$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Pb}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Pb}(\mathrm{s})$
the ratio $\frac{\left[\mathrm{Sn}^{2+}\right]}{\left[\mathrm{Pb}^{2+}\right]}$ when this cell attains equilibrium is
(Given $\mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{0}=-0.14 \mathrm{\;V}$ $\left.\mathrm{E}_{\mathrm{Pb}^{+2}/{\mathrm{Pb}}}^{0}=-0.13 \;\mathrm{V}, \frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06\right)$