If I take 25 ml of milk of magnesia (80 mg/ml of Mg(OH)2) and to this I slowly add 25 ml of 0.1 N HCl what will be the final pH?

Question

Milk of Magnesia has a documented pH of about 10.5 and indeed this is the pH I achieved experimentally when measuring with a pH meter. I have an excess of base 0.0171 moles vs 0.0025 moles of acid. I assume all the acid is neutralized leaving me with excess base. Does this mean that theoretically the pH of the final solution will revert to the pH of the base (10.5)? If nor, then what does happen, and can it be proved mathematically?Also, I had to mathematically prove why the pH of the milk of magnesia was 10.5. I used the Ksp of Mg(OH)2:

5.61 x 10 -12 = [Mg 2+][OH-]^2

5.61 x 10^-12 = [x] x [2x]^2 = 4x^3

x = 1.12 x 10^-4.  

[OH-]= 2x= 2.24 x 10^-4

pOH = - log [OH-]= 3.65

pH=10.35

is this correct?

 

Answer 1

Let's begin with your calculation (proof) of the pH for Mg(OH)₂.  You did do this correctly. Now, as for your other question.

as long as the concentration of the Mg(OH)2 exceeds the solubility, then yes, the pH will be that of the saturated solution, namely in this case, 10.35.