Question

If $pK _{ b }$ for $CN^-$ at $25^{\circ} C$ is 4.7 , the $pH$ of $0.5\, M$ aqueous $NaCN$ solution is:

• A. 12
• B. 10
• C. 11.5
• D. 11

Answer 1

Answer: (C)

$CN ^{-}+ H _{2} O \rightarrow HCN + OH ^{-} ; \,\,\,pK _{ b }=4.7$

$\therefore pK _{ a }=9.3$

$(\overset{-}{{O}} H )= ch = c \sqrt{\frac{ K _{ H }}{ c }}=\sqrt{ K _{ H } \cdot c }=\sqrt{\frac{ K _{ w }}{ K _{ a }} \cdot c }$

$=\sqrt{\frac{10^{-14} \times 0.5}{10^{-9.3}}}$

$\left[ OH ^{-}\right]=\sqrt{10^{-4.7} \times 0.5} $

$ \therefore pOH =11.5 $