Sebanyak 0.04 m larutan natrium sianida (ka HCN =4×10^-10 KW= 10^-14 hitunglah ph larutan garam tersebut

OH- = [tex] \sqrt{ \frac{Kw}{ka} . [G]} [/tex]
OH- = [tex] \sqrt{ \frac{ 10^{-14}}{4. 10^{-10}}.0.04 } [/tex]
OH- = [tex] 10^{-3} [/tex]

pOH = 3
pH = 14-3 = 11