Question

A weak acid, HA, is partially titrated using NaOH. You start with 50.0mL of a 0.100 M HA and add 30.0 mL of a 0.100 M NaOH, and measure the pH of the solution at 7.00. What is the pKA of Ha? What would be the pH of the titration at the equivalence point (when 50.0 mL of NaOH is added)?

I got the correct answer for the pKa (6.82), but don't know how to approach the second part. The answer to the second part ph pH = 9.76

Answer 1

We have, pKa = - log Ka = 6.82

$\therefore$K a = 10 ^{- pKa} = 10 ^{- 6.82} = 1.51 $\times$ 10 ^-07

Consider a reaction , HA + NaOH $\rightarrow$ NaA + H₂O

Now calculate equivalence point

We have relation, M acid $\times$ V acid = M base $\times$ V base

V base = M acid $\times$ V acid / M base

V base = 0.100 M $\times$ 50.0 ml / 0.100 M

V base =50.0 ml

Volume of NaOH needed to reach the equivalence point is 50.0 ml

At equivalence point all acid HA is consumed by added NaOH. pH of solution will be due to dissociation of NaA in water as

A ^-_(aq) + H₂O _(l) HA _(aq) + OH^-_(aq)  

For above reaction, K b = [HA] [ OH^-] / [A ^-] = K w / K a = 10 ^-14 / 1.51 $\times$ 10 ^-07 = 6.62 $\times$10 ^-08

mmol of NaA produced at equilibrium  = mmol of NaOH added = 0.100 $\times$ 50.0 = 5.00 mmol

Volume of solution at equivalence point = Volume of HA + Volume of NaOH = 50.0 + 50.0 = 100.0 ml

[NaA] = No. of moles of NaA / Volume of solution in L

[NaA] = [5.00 / 1000] / [100 /1000]

[NaA] =0.05 M

Let's use ICE table.

Concentration (M)	A -	HA	OH-
Initial	0.05
Change	-X	+X	+X
Equilibrium	0.05 - X	X	X

$\therefore$ K b = (X) (X) / 0.05 - X = 6.62 $\times$ 10 ^-08

X ² / 0.05 - X = 6.62$\times$ 10 ^-08

Assume X is very small as compared to 0.05. Then we can write 0.05 -X $\approx$ 0.05

$\therefore$ X ² / 0.05 = 6.62 $\times$ 10 ^-08

X ² = 0.05 $\times$ 6.62 $\times$ 10 ^-08

X ² = 3.31 $\times$ 10 ^-09

X = 5.75 $\times$ 10 ^-05 M = [HA] = [ OH^-]

We have relation, [H⁺] [ OH^-] = 10 ^-14

$\therefore$ [H⁺] = 10 ^-14 /   [ OH^-] = 10 ^-14 / 5.75 $\times$ 10 ^-05 = 1.74 $\times$ 10 ^-10 M

We have, pH = - log [H⁺] = -log 1.74 $\times$ 10 ^-10 = 9.76