0.4324 g sample of a potassium hydroxide-lithium hydroxide mixture requires 28.28 ml of 0.3520 M HCl for its titration to the equivalence point. What is the mass percent lithium hydroxide in this mixture?

Answer 1

You need two equations in two unknowns.
mass KOH + mass LiOH = 0.4324g
mols KOH + mols LiOH = 0.02828 x 0.3520
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Let X = mass KOH
and Y = mass LiOH
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X + Y = 0.4324
(X/molar mass KOH) + (Y/molar mass LiOH) = 0.02828 x 0.03520
Solve the two equations for X and Y, then,
mass % LiOH = (g LiOH/mass sample)*100 = ?
Post your work if you get stuck.

Answer 2

12.34%