Buffer Solution Chemistry Math Question 5

Question

80 mL of a .40 M solution of unknown weak acid is combined with 80 ml of a .25 M strong base, and the final pH of the solution is measured to be 4.13. What is the pK_a of the unknown acid?

Answer 1

HA = weak acid

OH- = strong base

HA + OH- ==> A- + H2O Where A- is now the conjugate base of HA. This forms a BUFFER solution.

moles HA = 0.080 L x 0.40 mol/L = 0.032 moles

moles OH- = 0.080 L x 0.25 mol/L = 0.02 moles

After reaction, you will have ....

moles HA = 0.032 - 0.02 = 0.012 moles HA

moles OH- = 0.02 - 0.02 = 0 moles OH-

moles A- = 0 + 0.02 = 0.02 moles A-

Total volume = 80 ml + 80 ml = 160 ml = 0.160 L

Final [HA] = 0.012 mol/0.160 L = 0.075 M

Final [A-] = 0.02 mol/0.160 L = 0.125 M

Using the Henderson Hasselbalch equation we can solve for the pKa of the acid:

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

4.13 = pKa + log (0.125/0.075) = pKa + log 1.67

4.13 = pKa + 0.22

pKa = 3.91