Calculation of pH (Chemistry)

Question

50.0mL of 0.40M of lactic acid is added with;

1.) 0.00mL

2.) 50.00mL

3.) 80.00mL

of .20M NaOH. Calculate the pH at each addition.

Answer 1

Lactic acid is a weak monoprotic acid. It's formula is CH3CH(OH)COOH. Since it is a weak acid, we need to know the Ka or pKa in order to answer this question. Looking it up on the internet, I find a pKa = 3.8

For the following, we can us HA to represent the weak acid.

HA <==> H⁺ + A^-

1) 0.40 M HA

Ka = [H⁺][A^-]/[HA] and Ka = 1x10^-3.8 = 1.58x10^-4

1.58x10^-4 = (x)(x) / 0.40 - x and if we assume x is small relative to 0.40 M, we can ignore it in the denominator

1.58x10^-4 = x² / 0.40

x² = 6.32x10^-5

x = 7.95x10^-3 M (and this is about 2% of the 0.4 M so our above assumption was valid)

7.95x10^-3 M = [H+] and pH = -log [H+]

pH = 2.10

2) 50.00 ml of 0.20 M NaOH is added. The NaOH reacts with the HA to produce H2O + NaA or H2O + A-

50.00 ml x 1 L/1000 ml x 0.20 mol/L = 0.01 moles NaOH added

Initial moles HA = 50.0 ml x 1 L/1000 ml x 0.40 mol/L = 0.02 moles HA

Final volume = 50.0 ml + 50.00 ml = 100.0 ml

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

0.02......0.01......................0........Initial

-0.01....-0.01....................+x........Change

0.01..................................0.01.....Equilibrium

Use Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid] = 3.8 + log [0.01/0.01]

pH = 3.8 = pKa

3) 80.00 ml of 0.20 M NaOH is added

80.00 ml x 1 L/1000 ml x 0.20 mol/L = 0.016 mol OH- added

50.00 ml x 1 L/1000 ml x 0.40 mol/L = 0.020 mol HA

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

0.02.......0.016......................0......Initial

-0.016...-0.016.................+0.016..Change

0.004........0......................0.016...Equilibrium

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

pH = 3.8 + log [0.016/0.004] = 3.8 + 0.6

pH = 4.4

Note: in calculations for #2 and #3, we don't need to use the final volume because we'd only be dividing by a constant to get molarity. Answer will come out the same.