The $\mathrm{pH}$ of an aqueous solution of $0.01 \mathrm{M} {\mathrm{CH}}_3{\mathrm{COONH}}_4$ at $25 °\mathrm{C}$ is:

$\left({\mathrm{K}}_{\mathrm{a}}\left({\mathrm{CH}}_3\mathrm{COOH}\right)={\mathrm{K}}_{\mathrm{b}}\left({\mathrm{NH}}_4\mathrm{OH}\right)=1.8\times {10}^{-5}\right)$

The dissociation constant of $\mathrm{HCN}$is $4.8\times {10}^{-10}$. What is the concentration of ${\mathrm{OH}}^{-}$ and $\mathrm{HCN}$ in a solution prepared by dissolving $0.16 \mathrm{mole}$ of $\mathrm{NaCN}$ in $450 \mathrm{mL}$ of water at $25°\mathrm{C}?$ 

The degree of dissociation of acetic acid in a $0.1 \mathrm{N}$ solution is $1.32\times {10}^{-2}$. At what concentration of nitrous acid will its degree of dissociation be the same as that of acetic acid?

${\mathrm{K}}_{\mathrm{a}} \mathrm{for} {\mathrm{HNO}}_2 = 4 \times {10}^{-4}$

Calculate the $\mathrm{pH}$ of the following aqueous solution(neglect the common ion effect).

$5\times {10}^{-8} \mathrm{M} \mathrm{HCl}$ 

(Use $\log \left(1.5\right)=0.17$)

Report your answer up to three significant figures.

Calculate the $\mathrm{pH}$ of the following aqueous solutions:

${10}^{-8} \mathrm{M} \mathrm{NaOH}$