Consider the titration of 600 mL of 0050 M CaOH2 solution wi

Consider the titration of 60.0 mL of 0.050 M Ca(OH)2 solution with 0.15 M HBr solution. A) What is the pH of the Ca(OH)2 solution before the start of the titration? B) 4. What is the pH of the solution after 10.0 mL of the HBr solution has been added to the 50.0 mL of 0.050 M Ca(OH)2 (a)? C) What is the pH of the solution which results at the equivalence point of the titration?

Solution

Ans. #A. Given,         [Ca(OH)₂] = 0.050 M

# 1 mol Ca(OH)₂ yields 2 mol OH^-.

So,

            [OH^-] in solution = 2 x [Ca(OH)₂] = 2 x 0.050 M = 0.100 M

Now,

            pOH = -log [OH^-] = -log 0.100 = 1.00

And,

            pH = 14.00 – pOH = 14.00 – 1.00 = 13.0

#B. Moles of Ca(OH)₂ = Molarity x Volume of solution in liters

                                                = 0.050 M x 0.050 L

                                                = 0.0025 mol

# 1 mol Ca(OH)₂ yields 2 mol OH^-.

So, total moles of OH^- in solution = 2 x 0.0025 mol = 0.0050 mol

# Moles of HBr added = 0.15 M x 0.010 L = 0.0015 mol

# 1 mol H⁺ from HBr neutralizes 1 mol OH^- from Ca(OH)₂.

So,

            Moles of OH^- neutralized = 0.0015 mol = moles of HBr added

            Remaining moles of OH^- = Initial moles – Moles neutralized

                                                = 0.0025 mol – 0.0015 mol

                                                = 0.001 mol

            Total volume of solution = 50.0 mL (base) + 10.0 mL (HBr) = 60.0 mL = 0.060 L

Now,

            Molarity of [OH^-] = Moles of OH^- / Volume of solution in liters

                                                = 0.001 mol / 0.060 L

                                                = 0.0167 M

            pOH = -log [OH^-] = -log 0.0167 = 1.78

            pH = 14.00 – 1.78 = 12.22

#C. Ca(OH)₂ is a strong base. HBr is a strong acid.

The pH at the equivalence point of strong acid- strong base titration is always equal to 7.00