A chemist titrates 220.0 mL of a 0.1917M propionic acid (HC2H, CO2) solution with 0.1787 M KOIH solution at 25 °C. Calculate the pH at equivalence. The pK of propionic acid is 4.89.

pH = 8.93

Explanation:

In this case, this titration is the case of a weak acid and a strong base. Now, at the equivalence point, it's supposed that we have the same moles of each reactant in solution, and we will expect that the pH would have to be 7. However, as the acid is pretty weak, there's a little difference in the solution because of the grade of dissociation of the acid, and the pH will be higher than 7. To know this, we first need to calculate the volume of added base:

M₁V₁ = M₂V₂

With this expression, let's calculate the volume of the base:

V₂ = M₁V₁ / M₂

V₂ = 0.1917 * 220 / 0.1787 = 236 mL

So, at the equivalence point, 236 mL are needed to neutralize this reaction. As the moles are the same for each reactant, we just need to calculate the concentration of the acid in this part. This will be the sum between the initial volume of acid and the calculated volume of base:

V of solution = 236 + 220 = 456 mL or 0.456 L

Then, the new concentration of the acid is:

[C₂H₅COOH] = 0.1917 * 0.220 / 0.456 = 0.0924 M

Now, the reaction with the base is the following:

C₂H₅COOH + KOH > C₂H₅COOK + H₂O

This means that in the equivalence point we have the propionic potassium and water, so, if take this and dissociates into it's ions we can calculate the pH of the solution:

C₂H₅COO⁻ + H₂O <> C₂H₅COOH + OH⁻

With this reaction in solution in the equivalence point, we just need the Kb of propionate ion, and this can be calculated with the value of the pKa of the acid:

Ka = 10^(-pKa)

Ka = 1.29x10⁻⁵

Now the value of Kb can calculated using the following expression:;

Kb = Kw / Ka ---> replacing we have

Kb = 1x10⁻¹⁴ / 1.29x10⁻⁵

Kb = 7.75x10⁻¹⁰

Now, with this value and the above reaction we can write an ICE chart to calculate the [OH⁻] and then, the pH of solution:

     C₂H₅COO⁻ + H₂O > C₂H₅COOH + OH⁻   Kb = 7.75x10⁻¹⁰

i)       0.0924                                    0               0

e)     0.0924-x                                  x               x

The Kb expression:

Kb = [C₂H₅COOH] [OH⁻] / [C₂H₅COO⁻]

7.75x10⁻¹⁰ = x² / 0.0924-x ---> Kb is very small, so this substraction can be neglected to just 0.0924 assuming x will be very small too.

7.75x10⁻¹⁰ = x² / 0.0924

7.75x10⁻¹⁰ * 0.0924 = x²

x = [OH⁻] = 8.46x10⁻⁶ M

With this value, we can calculate pOH and then the pH:

pOH = -log(8.46x10⁻¹⁰) = 5.07

Finally the pH:

pH = 14 - pOH

pH = 14 - 5.07