You are a CEM 262 TA and are preparing a 1.97 M ZnCl2 solution as an...

You are a CEM 262 TA and are preparing a 1.97 M ZnCl2 solution as an unknown for the EDTA titration experiment. The pH must be below what value to avoid precipitation of zinc hydroxide in the solution? The solubility product of Zn(OH)2 is 3×10-16. For this calculation, assume that the solution is ideal.

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Expert Solution

Ksp of Zn(OH)2 is 3 x 10^-16

A substance's solubility product, K_sp, is the mathematical product of its dissolved ion concentrations raised to the power of their stoichiometric coefficients. Solubility products are relevant when a sparingly soluble ionic compound releases ions into solution.

Which means when the product of the concentration of the dissolved ion concentrations raised to the power of their stoichiometric coefficients is below Ksp they will remain soluble.

1.97 M ZnCl2 will have 1.97 M Zn2+ and 2 x 1.97 M Cl^-in its solution

Zn(OH)₂ Zn²⁺ + 2 OH^-

Ksp = [Zn²⁺][OH^-]²

3 x 10^-16 = 1.97 x [OH^-]²

[OH^-]² = 3 x 10^-16/1.97

[OH^-]² = 1.52 x 10^-16

[OH^-] = 1.52 x 10^-16

[OH^-] = 1.23 x 10^-8M

so the concentration of [OH-] should be below 1.23 x 10^-8M for Zn(OH)₂ to be soluble

pOH = - log [OH^-]

pOH = - log (1.23 x 10^-8)

pOH = 7.91

So pOH should be more than 7.91

pH = 14 - pOH

pH = 14 - 7.91

pH = 6.09

So the pH must be below 6.09 for Zn(OH)₂ to be soluble

queen_honey_blossom answered 1 year ago

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