Question

A solution contains 1.60×10^-2 M zinc acetate and 1.60×10^-2 M calcium nitrate. Solid potassium carbonate is added slowly to this mixture. What is the concentration of zinc ion when calcium ion begins to precipitate? Solubility product constant data is found in the Chemistry References.

[Zn²⁺] = M

Answer 1

When calcium ion begins to precipitate as calcium carbonate

$[Ca^{2+}]=[Ca(NO_3)_2]=1.60 \times 10^{-2} \ M$

The solubility product constant for calcium carbonate is $K_{sp}=3.3 \times 10^{-9}$

The solubility product constant expression for calcium carbonate is $K_{sp}=[Ca^{2+}] \times [CO_3^{2-}]$

$3.3 \times 10^{-9} =1.60 \times 10^{-2} \ M \times [CO_3^{2-}]$

$[CO_3^{2-}]=\frac{3.3 \times 10^{-9} }{1.60 \times 10^{-2}}$

$[CO_3^{2-}]=\frac{3.3 \times 10^{-9} }{1.60 \times 10^{-2}}$

$[CO_3^{2-}]=2.06 \times 10^{-7} \ M$

When calcium ion begins to precipitate, the carbonate ion concentration is $2.06 \times 10^{-7} \ M$ . Use this information and solubility product constant of zinc carbonate to calculate zinc ion concentration.

The solubility product constant for zinc carbonate is $K_{sp}=1.0 \times 10^{-10}$

The solubility product constant expression for zinc carbonate is $K_{sp}=[Zn^{2+}] \times [CO_3^{2-}]$

$1.0 \times 10^{-10} =[Zn^{2+} ] \times 2.06 \times 10^{-7} \ M$

$[Zn^{2+} ] =\frac{1.0 \times 10^{-10}}{2.06 \times 10^{-7} }$

$[Zn^{2+} ] =4.85 \times 10^{-4} \ M$