Question

The solubility of $Pb(OH)_{2}$ in pure water is $\frac{20}{3}\times 10^{- 6}M$ . Calculate concentration of $OH^{-}$ ion in a buffer solution of $P^{H}=8$ .

• A. $\frac{20}{3}\times 10^{- 6}M$
• B. $\frac{10}{3}\times 10^{- 6}M$
• C. $10^{- 6}M$
• D. $2\times 10^{- 6}M$

Answer 1

Answer: (C)

As pH = 8

So $\left[\right.H^{+}\left]\right.=10^{- 8}$

$\left[\right.OH^{-}\left]\right.=\frac{K_{w}}{\left[\right. H^{+} \left]\right.}$

$\left[\right.OH^{-}\left]\right.=\frac{1 0^{- 14}}{1 0^{- 8}}=10^{- 6}$