Question

What is the final pH at 25°C when 56 mL of 0.139 mol/L NaOH(aq) and 30 mL of 0.187 mol/L HA(aq) are mixed. Assume that HA is a weak monoprotic acid with pKa = 4.99 at 25°C

Answer 1

Let us write down the acid-base reaction between the monoprotic acid and NaOH as

HA (aq) + OH^- (aq) ----------> A^- (aq) + H₂O

Now we calculate the moles of HA and NaOH available to us.

Amount of NaOH added = (56 mL/ 1000 mL).(1 L).(0.139 mol/L) = 0.00778 mole.

Amount of HA = (30 mL/1000 mL).(1 L).(0.187 mol/L) = 0.00561 mole.

As per the equation, NaOH reacts with HA in a 1:1 molar ration, hence the entire HA has been consumed and we have an excess of NaOH in the solution.

In other words, 0.00561 mole HA has reacted with 0.00561 mole NaOH and we have excess NaOH in the solution. We find out the excess NaOH in the solution as (0.00778 – 0.00561) = 2.17 x 10^-3. This amount of NaOH is present in the solution after the end-point. The volume of the solution = volume of HA present + volume of NaOH added = (30 + 56) = 86 mL.

The concentration of excess OH^- = (2.17 x 10^-3 mol/ 86 mL).(1000 Ml/ 1 L) = 0.0252 mol/L.

OH^- is a strong base and the pH will be decided by the concentration of OH^- (since it is in excess).

pOH = -log₁₀[OH^-] = -log₁₀(0.0252) = 1.598

Therefore, pH = 14 – pOH (since pH + pOH =14)

or, pH = 12.402

The final pH of the solution at 25°C is 12.402.