Chemistry Math Question?

Question

120.0 mL of a .40 M solution of unknown weak acid is combined with 90 ml of a .25 M strong base, and the final pH of the solution is measured to be 4.47. What is the pK_a of the unknown acid?

Answer 1

Let HA be the weak acid.

HA ==> H⁺ + A^-

moles HA present = 0.1200 L x 0.40 mole/L = 0.048 moles HA

moles strong base added = 0.090 L x 0.25 moles/L = 0.0225 moles base added

HA + OH^- ===> H2O + A^-

Assuming the reaction of base and acid is 1:1 on a mole basis, we have the following after addition of base....

0.048 moles HA - 0.0225 moles = 0.0255 moles HA left

0.0225 moles A- formed.

THIS FORMS A BUFFER i.e. a weak acid and the salt of the acid.

Using the Henderson Hasselbalch equation, we can find the pKa of the unknown acid:

pH = pKa + log [salt]/[acid]

In this case, the [salt] = [acid] = 0.0225 moles/0.21 L and so pH = pKa b/c log 1 = 0

pKa = 4.47