Question

generate a titration curve from the information provided: your titrand is 48.0 mL of 0.500 M lactic acid and you will be titrating with 1.0 M KOH.

Answer 1

We first get Initial pH:

CH₃CH(OH)CO₂H <-> H⁺ + CH₃CH(OH)CO₂^-

We do our ICE:

	CH3CH(OH)CO2H	<->	H+	+	CH3CH(OH)CO2-
I	0.5 M		0		0
C	-x		+x		+x
E	0.5 - x		x		x

Now we use the definition of constant:

1.38 x 10^-4 = [H⁺] [CH₃CH(OH)CO₂^-] / [CH₃CH(OH)CO₂H]

1.38 x 10^-4 = x² / (0.5 - x)

Isolating for x:

x = [H⁺] = 0.0082379 M

Now we get pH:

pH = -log (0.0082379) = 2.08

Now we get equivalence point

In which pH = pKa:

pH = pKa = -log (1.38 x 10^-4) = 3.86

Volume added = 0.012 mol / 1 M = 12 mL

At almost endpoint:

Initial moles: 0.048 L * 0.5 mol/L = 0.024 moles

Moles of base almost at endpoint = 0.023 moles

Volume of base = 0.023 moles / 1 mol/L = 23 mL

Concentrations:

[Base] = 0.023 mol / 0.071 L = 0.3239 M

[Acid] = 0.001 mol / 0.071 L = 0.014 M

pH = 3.86 + log (0.3239 / 0.014) = 5.2217

After endpoint:

pOH = -log(1) = 0

pH = 14 - 0 = 14

Titration curve: