Chapter 6: Q17 P (page 142)

(a) From the solubility product of zinc ferrocyanide, ${Zn}_{2} Fe {(CN)}_{6}$ , calculate the concentration of $Fe {(CN)}_{6}^{4 -}$ in 0.1 0m M ${ZnSO}_{4}$ saturated with ${Zn}_{2} Fe {(CN)}_{6}$ . Assume that ${Zn}_{2} Fe {(CN)}_{6}$ is a negligible source of ${Zn}^{2 +}$ .
(b) What concentration of $K_{4} Fe {(CN)}_{6}$ should be in a suspension of solid ${Zn}_{2} Fe {(CN)}_{6}$ in water to give $[{Zn}^{2 +}] = 5.0 \times 10^{- 7} M$ ?

Short Answer

Expert verified

The concentration of $Fe {(CN)}_{6}^{4 -}$ is $2.1 \times 10^{- 8} M$

The concentration of $K_{4} Fe {(CN)}_{6}$ is $8.4 \times 10^{- 4} M$

Step by step solution

Concept used

Solubility product $(K_{sp})$ :

It is defined as the product formed from the mathematical product of its concentration of dissolved ion raised to the power of its stoichiometric coefficient.

Calculate the concentration of Fe(CN)64-

${Zn}_{2} Fe {(CN)}_{6} ⇋ {Zn}^{2 +} + Fe {(CN)}_{6}$

The concentration is:

$K_{sp} = {[{Zn}^{2 +}]}^{2} [Fe {(CN)}_{6}^{4 -}] 2.1 \times 10^{- 16} = {(0.00010)}^{2} [Fe {(CN)}_{6}^{4 -}] [Fe {(CN)}_{6}^{4 -}] = \frac{2.1 \times 10^{- 16}}{10^{- 8}} [Fe {(CN)}_{6}^{4 -}] = 2.1 \times 10^{- 8} M$

Calculate the concentration of K4Fe(CN)6

$K_{sp} = {[{Zn}^{2 +}]}^{2} [Fe {(CN)}_{6}^{4 -}] 2.1 \times 10^{- 16} = {(5.0 \times 10^{- 7})}^{2} [K_{4} Fe {(CN)}_{6}^{4 -}] [K_{4} Fe {(CN)}_{6}^{4 -}] = \frac{2.1 \times 10^{- 16}}{{(5.0 \times 10^{- 7})}^{2}} [K_{4} Fe {(CN)}_{6}^{4 -}] = 8.4 \times 10^{- 4} M$