Question

59.2 mL of 1.98 M hydrochloric acid is added to 38.2 mL of barium hydroxide, and the resulting solution is found to be acidic.

16.5 mL of 5.53 M potassium hydroxide is required to reach neutrality.

What is the molarity of the original barium hydroxide solution?

Answer 1

excess 2HCl + Ba(OH)2 -----> BaCl2 + 2 H2O + unreacted HCl

unreacted HCl + KOH ----> KCl + H2O

Total moles of HCl = 59.2 mL x 1.98 mol / 1000 mL => 0.1172 moles

unreacted moles of HCl = 16.5 mL x 5.53 moles / 1000 mL => 0.0912 moles

So moles of HCl consumed in reaction with Ba(OH)2 = 0.1172 - 0.0912 => 0.026 moles

1 moles of Ba(OH)2 requires 2 moles of HCl

Therefore moles of Ba(OH)2 reacted with HCl = 0.026 / 2 => 0.013 moles

Molarity of Ba(OH)2 = 0.013 moles / 0.0382 L => 0.34 M