You have 75.0 mL of 0.10 M HA. After adding 30.0 mL of 0.10 M NaOH, the pH is 5.50. What is the Ka value of HA?

Question

You have 75.0 mL of 0.10 M HA. After adding 30.0 mL of 0.10 M NaOH, the pH is 5.50. What is the Ka value of HA?

the answer is:

2.1 x 10^-6

Please list steps so I can follow along. Thank you!

Answer 1

[HA] = 0.045x0.10/0.105 M

= (0.04286 millimoles)

[A^-] = 0.030x0.10/0.105 M

= (0.02857 millimoles)

The Henderson-Hasselbalch equation can be used:

pH = pK_a + log(A^-/HA)

5.50 = pK_a + log(0.02857/0.04286)

pK_a = 5.50+0.18 = 5.68

K_a =2.09x10^-6