Question

Determine the volume of 1.2 mol/L HCl that will be required to neutralize 150 mL of 0.080 mol/L Ba(OH)2

Answer 1

Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O

Mols Ba(OH)2 = M x L = approx 0.012
mols HC = twice that (look at the coefficients in the balanced equation.
HCl = appox 0.024
Then mol = M x L. You know M and mols, solve for L.