22. Considering the titration of a 25.0 mL sample of 0.115 NaOH with 0.100 M HCl,
calculate,
i) the initial pH (before adding any HCl solution).
ii the volume of acid required to reach the equivalence point.
iii) the pH after addition of 5.0 mL HCl.
iv) the pH at the equivalence point.

i) The initial pH before adding any HCl solution is 13.06

ii) The volume of acid required to reach equivalence point is 28.75 mL

iii) The pH after addition of 5.0 mL HCl is 12.899

iv) The pH at the equivalence point is 7

Explanation:

i) Here we have the concentration of the sample = 0.115M NaOH

Therefore pH of NaOH

pOH = -log[OH⁻] = -log[0.115] = 0.9393

pH + pOH = 14

∴ pH = 14 - pOH = 14 - 0.9393 = 13.06

The initial pH before adding any HCl solution = 13.06

ii) The equivalence point is the point of complete neutralization, therefore;

NaOH + HCl → NaCl + H₂O

One mole of NaOH reacts with one mole of HCl to form one mole of NaCl and one mole of H₂O

Therefore, 0.115 Mole of NaOH will be completely neutralized by 0.115 mole of HCl, hence we have

25 × 0.115 moles of NaOH ≡ x × 0.1 of HCl

Where:

x = Volume of HCl required in mL

∴ x = 25×0.115/0.1 = 28.75 mL of HCl is required to reach equivalence point

The volume of acid required to reach equivalence point = 28.75 mL

iii) The pH after addition of 5.0 mL HCl is found as follows

The volume, y of NaOH that is completely neutralized by 5.0 mL HCl is given by the following expression

y×0.115 = 5×0.1

y = 5×0.1/0.115 = 4.35 mL

Therefore, since the total volume is now 25 + 5 = 30 mL and the amount of 0.115 M NaOH in the solution is 25 - 4.35 = 20.65 mL, the molarity of the solution is therefore;

20.65 mL/1000 × 0.115 = 0.002375 moles in 30 mL

Number of moles in one liter = 0.002375/(30/1000) = 0.0792 moles

Hence the new molarity of NaOH = 0.0792 M

The pH = 14 - pOH = 14 - (-log[OH])

= 14 - (-log(0.0792)) = 12.899

The pH after addition of 5.0 mL HCl = 12.899

iv) The pH at equivalence point after complete neutralization is the pH of a neutral solution hence

pH at the equivalence point = 7.