Equal volumes of two solutions pH=2 and pH=4 are mixed together.Calculate the pH of the resulting solution?

^{+}] for solution with pH = 2 is 10

^{-2}M

[H

^{+}] for solution with pH = 4 is 10

^{-4}M

Since, equal volumes of the given solutions are mixed; concentration will be halved.

Therefore,

$\left[{H}^{+}\right]=\frac{{10}^{-2}+{10}^{-4}}{2}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{-2}(1+{10}^{-2})}{2}\phantom{\rule{0ex}{0ex}}=\frac{1.01x{10}^{-2}}{2}\phantom{\rule{0ex}{0ex}}=0.505x{10}^{-2}M$

$pH=-\mathrm{log}\left[{H}^{+}\right]\phantom{\rule{0ex}{0ex}}=-\mathrm{log}(0.505x{10}^{-2})\phantom{\rule{0ex}{0ex}}=2.29\phantom{\rule{0ex}{0ex}}=2.3$

**
**