Question

A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 40.0 mL ?

Answer 1

Consider a solution is made by mixing 8.00 mmol of HA and 3.00 mmol of the NaOH.

1) Calculate concentration of salt:

The 8.00 mmol of HA reacted in a 1:1 molar ratio with the NaOH to make 8.00 mmol of NaA.

8.00 mmol / 40.0 mL = 0.2 M

2) Calculate K_b of A¯:

(5.61 x 10^-6) (x) = 1.00 x 10^-14

x = 1.7825 x 10^-9

3) Calculate [OH¯]:

A¯ + H₂O <==> HA + OH¯

1.7825 x 10^-9 = [(x) (x)] / 0.2

x = 1.889 x 10^-5 M

4) Calculate pH

pOH = -log 1.864 x 10^-5 = 4.724

pH = 14.000 - 4.730 = 9.276