Question

A 25 mL aliquot of an HCl solution is titrated with 0.100 M NaOH. The equivalence point is reached after 21.27 mL of the base were added. Calculate 1) the concentration of the acid in the original solution, 2) the pH of the original HCl solution and the original NaOH solution, 3) the pH after 10.00 mL of NaOH have been added, 4) the pH at the equivalence point, and 5) the pH after 25.00 mL of NaOH have been added.

Answer 1

1) no. of mole = molarity X volume of solution in liter

no. of mole of NaOH = 0.100 X 0.02127 = 0.002127 mole

neutrilization reaction is

HCl + NaOH NaCl + H₂O

According to neutrilization reaction between HCl and NaOH they react equimolary therefore to react with 0.002127 mole of HCl react with 0.002127 mole of NaOH

no of mole of HCl = 0.002127 mole

volume of HCl = 25 ml = 0.025 L

molarity = no. of mole / volume of solution in liter

molarity of HCl = 0.002127 / 0.025 = 0.08508 M

molarity of HCl = 0.08508 M

2)

HCl is strong acid dissociate completly therefore [HCl] = [H⁺​]

pH = -log[H⁺] = -log(0.08508) = 1.07

pH of original HCl = 1.07

NaOH is strong base therefore [NaOH] = [OH^-]

pOH = -log[OH^-] = -log(0.100) = 1

pH = 14 - pOH = 14 - 1 = 13

pH of original NaOH solution = 13

3) no. of mole = molarity X volume of solution in liter

no. of moel of HCl = 0.025 X 0.08508 = 0.002127 mole

no. of mole of NaOH = 0.100 X 0.010 = 0.001 mole

neutrilization reaction is

HCl + NaOH NaCl + H₂O

According to neutrilization reaction between HCl and NaOH they react equimolary therefore to react with 0.001mole of HCl react with 0.001 mole of NaOH

no of mole of HCl remain in solution = 0.002127 - 0.001 = 0.001127 mole

total volume of solution = 25 + 10 = 35 ml = 0.035 liter

molarity = no. of mole / volume of solution

molarity of HCl = 0.001127 / 0.035 = 0.0322 M

HCl is strong acid dissociate completly therefore [HCl] = [H⁺​]

pH = -log[H⁺] = -log(0.0322) = 1.49

pH = 1.49

4) no. of mole = molarity X volume of solution in liter

no. of moel of HCl = 0.08508 X 0.025 = 0.002127 mole

no. of mole of NaOH = 0.100 X 0.02127 = 0.002127 mole

neutrilization reaction is

HCl + NaOH NaCl + H₂O

According to neutrilization reaction between HCl and NaOH they react equimolary therefore to react with 0.002127 moLe of HCl react with 0.002127 mole of NaOH

complete neutrilization take place and form salt and water and pH of solution = 7

5) no. of mole = molarity X volume of solution in liter

no. of moel of HCl = 0.08508 X 0.025 = 0.002127 mole

no. of mole of NaOH = 0.100 X 0.025 = 0.0025 mole

neutrilization reaction is

HCl + NaOH NaCl + H₂O

According to neutrilization reaction between HCl and NaOH they react equimolary therefore to react with 0.002127 moel of HCl react with 0.002127 mole of NaOH

no of mole of NaOH remain in solution = 0.0025 - 0.002127 = 0.000373 mole

total volume of solution = 25 + 25 = 50 ml = 0.050 liter

molarity = no. of mole / volume of solution

molarity of NaOH = 0.000373 / 0.050 = 0.00746 M

NaOH is strong base dissociate copletly therefore [NaOH] = [OH^-]

pOH = -log[OH^-] = -log(0.00746) = 2.13

pH = 14 - pOH = 14 - 2.13 = 11.87

pH = 11.87