Question

52.0 mL of 0.757 M hydrochloric acid is added to 12.1 mL of potassium hydroxide, and the resulting solution is found to be acidic.
20.8 mL of 0.630 M calcium hydroxide is required to reach neutrality.
what is the molarity of the original potassium hydroxide solution. __M

Answer 1

Balanced reaction 1 : HCl + KOH KCl + H2O

Balanced reaction 2 : Ca(OH)2 + 2 HCl CaCl2 + 2 H2O

concentration of HCl = 0.757 M

volume of HCl = 52.0 mL

moles of HCl added = (concentration of HCl) * (volume of HCl)

moles of HCl added = (0.757 M) * (52.0 mL)

moles of HCl added = 39.364 mmol

concentration of Ca(OH)2 = 0.630 M

volume of Ca(OH)2 = 20.8 mL

moles of Ca(OH)2 added = (concentration of Ca(OH)2) * (volume of Ca(OH)2)

moles of Ca(OH)2 added = (0.630 M) * (20.8 mL)

moles of Ca(OH)2 added = 13.104 mmol

moles of HCl consumed by Ca(OH)2 = (moles of Ca(OH)2 added) * (2 moles HCl / 1 mole Ca(OH)2)

moles of HCl consumed by Ca(OH)2 = (13.104 mmol) * (2 / 1)

moles of HCl consumed by Ca(OH)2 = (13.104 mmol) * 2

moles of HCl consumed by Ca(OH)2 = 26.208 mmol

moles of HCl consumed by KOH = (moles of HCl added) - (moles of HCl consumed by Ca(OH)2)

moles of HCl consumed by KOH = (39.364 mmol) - (26.208 mmol)

moles of HCl consumed by KOH = 13.156 mmol

moles of KOH present = moles of HCl consumed by KOH

moles of KOH present = 13.156 mmol

molarity of KOH = (moles of KOH present) / (volume of KOH solution)

molarity of KOH = (13.156 mmol) / (12.1 mL)

molarity of KOH = 1.09 M