Question

If $p{K}_b$ for $C{N}^{-}$ at ${25}^o$C is $4.7$. The $pH$ of $0.5M$ aqueous $NaCN$ solution is?

• A. $12$
• B. $10$
• C. $11.5$
• D. $11$

Answer 1

The correct option is C $11.5$

$C{N}^{-}$ will come from complete ionization of NaCN, Thus

Reaction can be represented as:

$C{N}^{-}+H_{2}O\to HCN+O{H}^{-}$

initial concentrations:

$0.500$

at equillibrium:

$0.5-xxx$

given that $p{K}_b=4.7$

as $p{K}_b=-log{K}_b$

$4.7=-log{K}_b$

or $K_b=2\times {10}^{-5}$

since $K_b=\frac{\left[HCN\right]\left[O{H}^{-}\right]}{\left[C{N}^{-}\right]}$

$2\times {10}^{-5}=\frac{x^2}{0.5-x}$

x is very small thus $0.5-x\approx 0.5$

$x=\left[O{H}^{-}\right]=3.16\times {10}^{-3}$

pOH=-log([$O{H}^{-}$])

pOH=-log ($3.16\times {10}^{-3}$)

pOH=2.5

using $pOH+pH=14$

we get $pH=14-2.5=11.5$

thus pH of the solution is 11.5.