How many moles of solid Mg(OH)2 should you add to 200mL of 0.15M lactic acid solution to produce pH 4.0 buffer?

Answer 1

Look up the pKa for lactic acid.
You have 200 x 0.15M HL(lactic acid) = 30 millimols HL.

Set up an ICE chart. You have 30 mmols HL and zero OH^- and zero L^-. You want to add NaOH to it to form the L^- and take away from the HL while doing that.
...........HL + OH^- ==> L^- + H2O
I.........30.....0.......0......0
add..............x..................
C.........-x....-x.......x........
E........30-x....0.......x

Now substitute the E line into the HH equation.
4.0 = pKa + log (x)/(30-x)
Solve for x = millimols OH^- needed to do the job. I would convert that to mols, take half of it (since Mg(OH)2 provides twice as much OH per 1 mol Mg(OH)2), then mols x molar mass = grams Mg(OH)2.
I never work one of these without checking it and I recommend you do the same. Start with ? grams Mg(OH)2 [I think the g is approximately 0.5 g], add that to the 0.030 moles HL, check to see how much HL is left and how much L^- is formed (you can do that with regular stoichiometry), stick those back into the HH equation and see if you end up with a pH of 4.0. You did something wrong if you don't get that; I did when I checked my answer.
Post your work if you get stuck.