Question

Suppose 0.127g of zinc chloride is dissolved in 50.mL of a 23.0mM aqueous solution of potassium carbonate. Calculate the final molarity of zinc cation in the solution. You can assume the volume of the solution doesn't change when the zinc chloride is dissolved in it. Be sure your answer has the correct number of significant digits.

Answer 1

Mass of Zinc chloride=0.127 g

Molar mass of Zinc chloride=molar mass of Zn+2xmolar mass of Cl=65.4 g/mol+2x35.5 g/mol=65.4 g/mol+71 g/mol=136.4 g/mol

Number of moles of ZnCl₂ added=mass/molar mass=0.127 g/136.4 g/mol=0.00093 mol

Number of moles of potassium carbonate=Molarity x volume (L)

=23.0 mM x 50 mL/1000 mL/L

=23.0 x 10^-3 M x 0.050 L= 0.00115 mol (1 L=1000 L and 1 mM=10^-3 M)

Zinc chloride reacts with potassium carbonate to form zinc carbonate (s) and potassium chloride (aq) (as solubility of zinc carbonate in water is very low it precipitates out)

1 mol K₂CO₃ reacts with 1 mol ZnCl₂​​ to form 1 mol ZnCO₃(s)

0.00115 mol K₂CO₃ will require 0.00115 mol ZnCl₂ but we have only 0.00093 mol ZnCl₂ (which is therefore the limiting reactant)

So 0.00093 ZnCl₂ reacts with 0.00093 mol K₂CO₃ to form 0.00093 mol ZnCO₃

Remaining number of moles of K₂CO_​​​3=0.00115 mol-0.00093 mol=0.00022 mol

As 1 mol potassium carbonate gives 1 mol carbonate ions in solution, so number of moles of CO₃^2- remaining in solution=0.00022 mol

Concentration of carbonate ions=number of moles/volume (L)

=0.00022 mol/0.050 L=0.0044 M

Ksp of ZnCO ₃=1.46x10^-10

Dissolution of ZnCO₃ in solution is as shown

s s

Where s is molar solubility of ZnCO₃

Ksp=[Zn^{​​​​​2+}][CO₃^2-]

1.46x10^-10=s x (s+0.0044 M) (carbonate ions are also present in solution due to unreacted K₂CO₃)

(As Ksp of ZnCO₃ is very small so s<<<0.0044 M and can be ignored in the second bracket)

1.46x10^-10=sx0.0044

s=1.46x10^-10/0.0044 =3.3 x10-8 M

So concentration of zinc ions in given solution=3.3x10^-8 M