Question

Suppose that 26 mL of 0.10 M HCl is added to 1.000 L of saturated Mg(OH)2' which contains more than enough Mg(OH)2(s) to react with all of the HCl.

A.) After the reaction has ceased, what will the molar concentration of Mg2+ be?

B.) What will the pH of the solution be?

Answer 1

a) Mg(OH)2 + 2 HCl --> MgCl2 + 2 H2O
Adding the HCl with excess Mg(OH)2 is equivalent to simply dissolving MgCl2 first, and then letting the excess Mg(OH)2 saturate the solution. To start,
(0.026 L) (0.10 mole/L) = 0.0026 moles
of HCl were added. Since the final volume is
1.000 L + 0.026 L = 1.026 L, then
[MgCl2] = (1/2) (0.0026 moles) / (1.026 L) = [Mg++] = 0.001267 M.
Note the (1/2) is from the stoichiometry of the acid-base reaction. The additional Mg++ from the excess Mg(OH)2 will be negligible compared to this.

b) Mg(OH)2 <--> Mg++ + 2 OH-  

let( standard) Ksp value of  Mg(OH)2 is =7.1x10-12.
Ksp = [Mg++] [OH-]²
Let x = molar solubility of Mg(OH)2, so
Ksp = (x) (2x)²
Since we have just calculated [Mg++]o above, and substituting for Ksp, we have:
7.1x10^-12 = (0.001267 + x) (2x)²
To solve this, we assume 0.001267>>x, so we ignore it in the left parentheses:
7.1*10^-12 = (0.001267) (2x)²
Solving, x = 3.7428x10^-5 M.
(We have just shown that the additional [Mg++] is indeed negligible compared to [Mg++]o.) We see from the stoichiometry
[OH-] = 2 [Mg++] = 7.4856x10^-5 M.
pOH = -log[OH-] = 4.12576
pH = 14 - pOH = 9.8742.