0.4276-g sample of a potassium hydroxide – lithium hydroxide mixture requires 31.20 mL of 0.3315 M HCl for its titration to the equivalence point. What is the mass percent lithium hydroxide in this mixture?

Question

1.) 0.03120 L x 0.3315= 0.010343 moles HCl
I'm not sure where I go from there..

Answer 1

This is not an easy problem. You set up two equations and solve them simultaneously. I will give you the two equations and tell you what to do after that.
Let X = grams LiOH
and Y = grams KOH
equation 1 is X+Y=0.4276

Equation 2 is obtained by an equation that adds mols LiOH (in terms of X) + mols KOH (in terms of Y) = total mols HCl I will use mm for molar mass.
(X/mm LiOH) + (Y/mm KOH) = total mols HCl = M HCl x L HCl.
Solve these equations simultaneously for X (and Y if you want it).
Then %LiOH = (grams LiOH/mass sample)*100 = ? Note: grams LiOH is X in your answer. Post your work if you get stuck.

Answer 2

@DrBob22 thank you for explaining. But how would you solve for x?

(x/23.947)+(y/56.105)=0.01034
(X/23.947))=0.01034-(y/56.105)...

Answer 3

Your problem isn't chemistry at this point; it's math. You haven't used the first equation of X+Y = 0.4276. Solve that for Y as
Y = 0.4276-X and substitute that into the equation you have for Y. That will give you just one unknown of X and you solve for that.