Question

•
Find
out
the
pH
of the solution
at 25°c
when [H+] = 2.0 x 10-3 M
When [OH-] = 10 x 10-5 M​

Answer 1

Explanation:

When [H+] = 2.0 × 10^(-3)

$pH = - log_{10} [ {H}^{+}] \\ \\ Where \: \: [ \: ] \: is \: concentration$

$pH = -log_{10} [2 × {10}^{-3}] \\ \\ pH = 3 - 0.3 \\ \\ \boxed{pH = 2.7}$

When [OH-] = 10 × 10^(-5) = 10^(-4)

$p</strong><strong>O</strong><strong>H = - log_{10} [ {</strong><strong>O</strong><strong>H}^{</strong><strong>-</strong><strong>}] \\ \\ Where \: \: [ \: ] \: is \: concentration$

$p</strong><strong>O</strong><strong>H = -log_{10} [ {10}^{</strong><strong>-</strong><strong>4</strong><strong>}</strong><strong>] </strong><strong>\\ \\</strong><strong> </strong><strong>p</strong><strong>O</strong><strong>H = </strong><strong>4</strong><strong> \\ \\ \boxed{pH =</strong><strong> </strong><strong>1</strong><strong>4</strong><strong> </strong><strong>-</strong><strong> </strong><strong>4</strong><strong> </strong><strong>=</strong><strong> </strong><strong>1</strong><strong>0</strong><strong>}$