Question

Calculate [OH−] for each solution. pH = 1.52, pH = 13.38, pH = 8.09, pH = 2.32

Answer 1

pH + pOH = 14

$\begin{gathered} \text{pOH =-log}_{10}\lbrack OH^-\rbrack \\ \lbrack OH^-\rbrack=10^{-(14-pH)} \end{gathered}$

For pH = 1.52

[OH-] = 3.311x10^-13 M

For pH 13.38

[OH-] =0.2399 M

For pH=8.09

[OH-] = 1.230x10^-6 M

For pH 2.32

[[OH-] = 2.089x^-12 M