For the following, consider Mg(OH) 2 (s) dissolved in 1.00 M NH 4 Cl(aq). Mg(OH)

Question

For the following, consider Mg(OH)2(s) dissolved in 1.00 M NH4Cl(aq).

Mg(OH)2(s) <----> Mg2+(aq) + 2 OH-(aq) Ksp = 1.8x10-11

NH3(aq) + H2O(l) <------> NH4+(aq) + OH-(aq) Ksp = 1.8x10-5

A) What is the numerical value of the equilibrium constant for the neutralization that occurs when Mg(OH)2 is added to the solution?

B) Find the equilibrium concentrations for the equilibrium of part (A) in terms of x, the change in initial concentrations. Plug these values into the equilibrium constant expression of part (A), and rearrange the terms to make a cubic expression in the form of ax3 + bx2 + cx + d = 0.

C) The cubic equation of part (B) has 2 imaginary roots and one real root. Given the real root is x = 0.178, find the solubility of Mg(OH)2 in this ammonium chloride solution in g/L.

Explanation / Answer

NH4+ + H2O = NH3 + H3O+
Ka = [NH3] [H+] / [NH4+]

NH3 + H2O = NH4+ + OH-
Kb = Kw / Ka = 1.8e-5 = [NH4+] [OH-] / [NH3]

Ksp = [Mg++] [OH-]^2
Keq = Ksp / Kb^2 = Ksp (Ka^2 / Kw^2) = [Mg++] [NH3]^2 / [NH4+]^2

Then you calculated s = [Ca++] in 1M NH4+:
Keq = [Mg++] [NH3]^2 / [NH4+]^2 = s (2s)^2 / (1M - 2s)^2
which is a cubic, not a quadratic (so we're not thinking along the same lines, you and I). However, if
s << 0.5M,
then the term in the denominator is still about 1M, and
4s^3 ~ Keq
s ~ 0.191 M
which is not nearly as small as I'd hoped, so I'll refine the result with a fixed-point interation:
s'^3 = Keq (1M - 2s)^2 / 4
where s' is the new, better estimate and s is the previous estimate. Here are my estimates:
s1=0.1911, s2=0.1386, s3=0.1539, s4=0.1495 (=s3 to 2 sig figs; done)

So, s = [Ca++] = 0.15M, [NH3] = 2s = 0.30M, [NH4+] = 1M - 2s = 0.7M, [OH-] = sqrt(Ksp / s) = 7.7e-6M, [H+] = Kw / [OH-] = 1.3e-9

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