A 0.4324 g sample of a potassium hydroxide - lithium hydroxide mixture requires 28.28 mL of 0.3520 M HC1 for its titration to the equivalence point . What is the mass percent lithium hydroxide in this mixture ?
Two equations and solve simultaneously.
Let X = grams KOH
Let Y = grams LiOH
mm = molar mass
am = atomic mass
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eqn 1..... X + Y = 0.4324
eqn 2..... mols KOH + mols LiOH = mols HCl = M HCl x L HCl = 0.3520 x 0.02828 = 0.009924
--------------------------------------------------------------------------------------------------eeqn 1....X + Y = 0.4324
eqn 2....(X/mmKOH) + (Y/mmLiOH) = 0.009924
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eqn 1......X + Y = 0.4324
eqn 2......(X/56.1) + (Y/23.9) = 0.009924
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Solve the two equations for Y = grams LiOH and plug into the following to obtain percent LiOH.
mass % LiOH = (grams LiOH/mass sample)*100 = ?
The chemistry part is above. The rest of this is just math, Post your work if you get stuck. NOTE: For the above equation 2 I have used molar masses (mm) rounded to the nearest tenth but you have four significant figures in the problem; therefore, you should recalculate molar masses of KOH and LiOH to four significant figures for a more accurate answer to the problem.
As I said above I've set up the chemistry for you. The rest is grind it out math.