Question

A student used the following amounts of reactants: 16.2431 g AcOH (which was 15.7 mL), 10.0822 g CH₃OH (which was 12.7 mL), and 0.1204 g conc. H₂SO₄ (which was two drops). After one week, the mixture was titrated with a 1.20 M NaOH solution to a phenolphthalein end point with 78.63 mL of the NaOH solution. What is the actual Kc?

Answer 1

Concentration of AcOH = moles/volume (L)

moles of AcOH = 16.2431g/60 g/mol = 0.270 moles, volume = 15.7mL = 0.0157 L

[AcOH] = 0.270/0.0157 = 17.19 mol/L

moles of CH3OH = 10.0822g/32g/mol = 0.315 moles, volume = 12.7ml = 0.0127 L

[CH3OH] = 0.315/0.0127 = 24.80 mol/L

The reaction between AcOH and NaOH is given by

CH3COOH + NaOH ----> CH3COONa + H2O

1mole CH3COOH = 1mole NaOH

The concentration of AcOH remains after one week can be found as

Millimoles of NaOH consumed = 1.20 M x 78.63 mL = 94.35 mmol = 0.09435 mol/L

[NaOH] reacted = [AcOH] remains = 0.09435 mo/L

[AcOH] reacted = 17.19 - 0.09435 = 17.09 mol/L

The reaction between AcOH and CH3OH is given by

CH3COOH + CH3OH === CH3COOCH3 + H2O

1mole CH3COOH = 1 mole CH3OH = 1mole CH3COOCH3

[CH3OH]remains = initial - consumed = 24.80 - 17.09 = 7.71 mol/L

[CH3COOCH3] formed = 17.09 mol/L

[H2O] can be neglected

Thus, Kc = [products]/[reactants]

Kc = (17.09)/(0.09435)(7.71) = 23.5