Question

Calculate the final pH of a solution made by the addition of 10 mL of a 0.5 M NaOH solution to 500 mL of a weak acid, HA. The concentration of the conjugate acid is 0.2M, the pH is 5.00 and thepKa = 5.00. Neglect the change in volume. (This is all of the information that was given.)

a.6.10

b. 5.04

c. 7.00

d. 5.99

e. 6.91

Answer 1

We have 500mL of a solution of weak acid, HA.

The pH of the solution = 5.00

pKa = 5.00

Conjugate acid concentration [HA] = 0.2M

From Henderson–Hasselbalch equation:

pH = pKa + log[A-]/[HA]

5= 5 + log[A-]/[HA]

Therefore, [HA] = [A-] = 0.2 M

To the above solution, we are adding 10mL of 0.5M NaOH

No. of moles of OH- = 0.5M x 0.01 L = 0.005 moles

0.005 moles of NaOH reaction with HA to form A-.

Therefore, after addition:

No. moles of HA= 0.2M x 0.5 L – 0.005 = 0.095 moles

[HA] = no.moles/volume in L = 0.095 moles/0.5 L = 0.19 M

(note: we are neglecting the volume change as mentioned in the problem)

No. moles of A-= 0.2 x 0.5 L +0.005 = 0.105 moles

[A-] = no.moles/volume in L = 0.105 moles/0.5 L = 0.21 M

Therefore, pH of the solution is :

pH = pKa + log[A-]/[HA]

= 5.00 + log( 0.21)/( 0.19) = 5.04

The pH of the solution = 5.04, therefore, answer is “b”