What volume of .300M HCL is required (a) to titrate 50.0 ml of 0.600 M LiOH? (B) to titrate 50.0ml of .600M Ca(OH)2

Question

(balanced equation and solution for both please!)

Answer 1

I'll do the Ca(OH)2 but the process is the same for LiOH.

Ca(OH)2 + 2HCl ==> 2H2O + CaCl2

step 1. Write and balance the equation.

step 2. Convert what you have, in this case Ca(OH)2), to mols. mols = M x L = 0.600 x 0.050 = 0.03 mols Ca(OH)2.

step 3. Using the coefficients in the balanced equation, convert mols of what you have [Ca(OH)2] to mols of what you want [HCl].
0.03 mols Ca(OH)2 x [2 mol HCl/1 mol Ca(OH)2] = 0.03 x 2/1 = 0.06 mols HCl. [Note how the factor convert mols Ca(OH)2 to mols HCl by canceling unit of mols Ca(OH)2 and but leaving unit of HCl].

step 4. Now you have mols HCl.
M HCl = mols HCl/L HCl. You have 0.3M HCl and mols HCl = 0.06, solve for L.
L = mols/M = 0.06/0.3 = 0.2L = 200 mL.