Step 1

HClO_{4} → H^{+ }+ ClO^{-}

$\frac{\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{HClO}}}{\mathbf{1}\mathbf{}\mathbf{L}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{H}}^{\mathbf{+}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{HClO}}}$** = 3.4x10 ^{-8} mol H^{+}**

What is the pH of a 3.4 x 10^{-8} M HClO_{4} solution at room temperature?

A. 7.1

B. 7.0

C. 5.3

D. 6.9