Question

Consider the titration of a 34.0 mL sample of 0.170 M HBr with 0.200 M KOH. Determine each of the following:

1. initial pH

2. the volume of added base required to reach the equivalence point

3. pH at 10.4 mL of added base

4. pH at the equivalence point

Answer 1

1.

HBr is a strong acid. Hence, 0.170 M HBr gives 0.170 M of H₃O⁺ in aqueous solution.

Thus,

pH = - log[H₃O⁺]

     = - log(0.170)

     = 0.770

Hence, the initial pH of the solution = 0.770

2.

Volume of acid (V₁) = 34.0 mL

Concentration of acid (S₁) = 0.170 M

Let us say that the volume of base is required = V₂ mL

Concentration of base (S₂) = 0.200 M

Now, at equivalence point,

moles of acid = moles of base

or, V₁ x S₁ = V₂ x S₂

or, 34.0 mL x 0.170 M = V₂ x 0.200 M

or, V₂ = 28.9 mL

Hence, the volume of base required to reach the equivalence point = 28.9 mL

3.

34.0 mL of 0.170 M HBr = 34.0 mL x 0.170 M = 5.78 mmol of HBr

10.4 mL of 0.200 M of KOH = 10.4 x 0.200 M = 2.08 mmol of KOH

Now, 2.08 mmol of KOH neutralizes 2.08 mmol of HBr.

HBr + KOH KBr + H₂O

Hence, the moles of unreacted HBr in the solution = (5.78 - 2.08) mmol = 3.70 mmol of HBr

Total volume of the solution after mixing = (34.0 + 10.4) mL = 44.4 mL

Molarity of unreacted KOH in the solution = 3.70 mmol/44.4 mL = 0.0833 M

Thus,

pH = - log[H₃O⁺]

     = - log(0.0833)

     = 1.08

4.

The pH at the equivalence point for a titration between a strong acid HBr and a strong base KOH = 7