The solubility of Pb(OH)2 in pure water is (20/3)× 10- 6M . Calculate concentration of OH- ion in a buffer solution of PH=8 .

Question

The solubility of Pb(OH)2 in pure water is (20/3)× 10- 6M . Calculate concentration of OH- ion in a buffer solution of PH=8 . (A) (20/3)× 10- 6M (B) (10/3)× 10- 6M (C) 10- 6M (D) 2× 10- 6M

Tardigrade Admin · Accepted Answer

As pH = 8 So [.H+].=10- 8 [.OH-].=(Kw/[. H+ ].) [.OH-].=(1 0- 14/1 0- 8)=10- 6

Q. The solubility of $P b (O H)_{2}$ in pure water is $\frac{20}{3} \times 1 0^{- 6} M$ . Calculate concentration of $O H^{-}$ ion in a buffer solution of $P^{H} = 8$ .

$\frac{20}{3} \times 1 0^{- 6} M$

$\frac{10}{3} \times 1 0^{- 6} M$

$1 0^{- 6} M$

$2 \times 1 0^{- 6} M$

As pH = 8

So $[H^{+}] = 1 0^{- 8}$

$[O H^{-}] = \frac{K _{w}}{[ H ^{+} ]}$

$[O H^{-}] = \frac{1 0 ^{- 14}}{1 0 ^{- 8}} = 1 0^{- 6}$

Q. The solubility of Pb(OH)2​ in pure water is 320​×10−6M . Calculate concentration of OH− ion in a buffer solution of PH=8 .

320​×10−6M

310​×10−6M

10−6M

2×10−6M