Question

In the titration of a solution of weak monoprotic acid with a 0.1275 M solution of NaOH, the pH half way to the equivalence point was 4.48. In the titration of a second solution of the same acid, exactly twice as much of a 0.1275 M solution of NaOH was needed to reach the equivalence point. What was the pH half way to the equivalence point in this titration?

Answer 1

2. The reaction between monoprotic acid and NaOH is

HA + NaOH -------> NaA + H2O

At half equivalence point,

the number of moles of NaOH remaining is equal to the number of moles of NaA formed.

So conc. of HA and Conc. of NaA are equal.

So pH = pKa + log{[NaA]/[HA]} = pKa + log 1 = pKa

In the first case, pH at half equivalence point is 4.48

So, pKa of the mono-protic acid = 4.48

Second case, whatever may be the concentration of NaOH, pH at half way point is 4.48

pH of the solution = 4.48