Question

31.4 mL of 0.657 M nitric acid is added to 46.5 mL of calcium hydroxide, and the resulting solution is found to be acidic.

20.9 mL of 0.742 M potassium hydroxide is required to reach neutrality.

What is the molarity of the original calcium hydroxide solution?

Answer 1

The balanced reaction is

Ca(OH)2 (aq) + 2HNO3 (aq) <---> Ca(NO3)2 (aq) + 2H2O (l)

Moles of nitric acid used = M x V ( in L)   = 0.657 x 0.0314 = 0.02063

Moles of KOH = 0.742 x 0.0209 = 0.0155

Nitric acid moles total used = nitric acid moles used to neutralise Ca(OH)2 + nitric acid moles ued to

                                            neutralsie KOH

0.02063 = Nitirc acid used for Ca(OH)2 + 0.0155

Nitric acid moles used for Ca(OH)2 = 0.00512

Ca(OH)2 moles used = HNO3 moles /2   ( as per coeffients of balanced equation)

                   = 0.00512 /2 = 0.00256

volume of Ca(OH)2 used = 0.0465 L

Molarity = moles / volume = 0.00256 /0.0465 =0.055 M

Thus Calcium hydroxide molarity = 0.055 M