Question

Solid lead nitrate is slowly added to 125 mL of a 0.0598 M sodium chromate solution. The concentration of lead ion required to just initiate precipitation is

Answer 1

$Pb CrO_4 \rightleftharpoons Pb^{2+}+CrO_4^{2-}$

$K_{sp}=[ Pb^{2+}] \times [CrO_4^{2-}]$ ... ....(1)

$K_{sp}=2.0 \times 10^{-16}$

$[CrO_4^{2-}]=[Na_2CrO_4]=0.0598 \ M$

Substitute values in equation (1)

$K_{sp}=[ Pb^{2+}] \times [CrO_4^{2-}]$

$2.0 \times 10^{-16}=[ Pb^{2+}] \times 0.0598 \ M$

$[ Pb^{2+}]=3.34 \times 10^{-15} \ M$