Question

Complete the table for each of the following solutions:

	[H+]	pH	pOH	[OH-]
0.0070 M
3.0 M KOH

Short answer

	[H+]	pH	pOH	[OH-]
0.0070 M HNO3	0.0070 M	2.15	1.43×10-12	11.85
3.0 M KOH	3.3×10-15	14.5	-0.5	3 M

Step-by-step solution

Calculation of various species for HNO3

HNO₃ is a strong acid, so [H⁺] = 0.0070 M

${\text{pH=\hspace{0.17em}-log[H}}^{\text{+}}\text{]=-log(0.0070)=2.15}$

$\left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\text{H}}^{+}\right]}=\frac{1\times {10}^{-14}}{7\times {10}^{-3}}=1.43\times {10}^{-12}$

$\text{pOH=\hspace{0.17em}14-pH=14-2.15=11.85}$

Calculation of various species for KOH

KOH is a strong base, so [OH^-] = 3 M

$\left[{\text{H}}^{+}\right]\text{=}\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[\mathrm{O}{\text{H}}^{-}\right]}=\frac{1\times {10}^{-14}}{3}=3.3\times {10}^{-15}$

${\text{pH=\hspace{0.17em}-log[H}}^{\text{+}}\text{]=-log(}3.3\times {10}^{-15}\text{)=14.5}$

$\text{pOH=\hspace{0.17em}14-pH=14-14.5=-0.5}$