Determine the pH

Question

Determine the pH at the point in the titration of 40.0 mL of 0.200 M HC4H7O2 with 0.100 M Sr(OH)2 after 10.0 mL of the strong base has been added. The values of Ka for HC4H7O2 is 1.5x10^-5.

Before (mol)

Change (mol)

After (mol)

Answer 1

2HC₄H₇O₂ + Sr(OH)₂ ==> 2H₂O + Sr(C4H7O2)₂

Initial mols HC₄H₇O₂ = 40.0 ml x 1 L /1000 ml x 0.200 mol/L = 0.008 moles HC₄H₇O₂

Initial mols Sr(OH)₂ = 10.0 ml x 1 L / 1000 ml x 0.100 mol/L = 0.001 mols Sr(OH)₂ = 0.002 mols OH-

Using an ICE table:

2HC₄H₇O₂ + 2OH^- ==> 2H₂O + 2C₄H₇O₂^-

0.008...........0.002.................................0..............Initial

-0.002...........-0.002.............................+0.002........Change

0.006..............0.....................................0.002........Equilibrium

This now has formed a buffer containing the weak acid HC₄H₇O₂ and the conjugate base, C₄H₇O₂^-

Use the Henderson Hasselbalch equation to find the pH:

pH = pKa + log [conj.base] /[acid]

pKa = -log 1.5x10^-5 = 4.82

pH = 4.82 + log (0.002/0.006) = 4.82 + (-0.48)

pH = 4.34