Question

A sample of 0.1387 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.1150 MNaOH. The acid required 15.5 mL of base to reach the equivalence point.

Part B After 7.25 mL of base had been added in the titration, the pH was found to be 2.45. What is the Ka for the unknown acid? (Do not ignore the change, x, in the equation for Ka.)

Answer 1

if monoprotic

mmol of acid = mmol of base

mmol of base = Mbase*Vbase = 15.5*0.1150 = 1.7825 mmol of base

mmol of acid = 1.7825

[Acid] = mmol/V = 1.7825/25 = 0.0713 M of acid

Molar MAss of acid = mass of acid / moles of acid = (0.1387)/(1.7825*10^-3) = 77.81206 g/mol

B)

V = 7.25 mL of base added.. pH = 2.45

find Ka

pH = pKa + log(A-/HA)

2.45 = pKa + log(A-/HA)

initially

mmol of HA = 1.7825

mmol of A- = 0

after mmol of base = MV = 7.25 *0.1150 = 0.83375 mmol of base is added

mmol of HA = 1.7825-0.83375 = 0.94875 mmol of acid left

mmol of A- = 0 + 0.83375 formed

2.45 = pKa + log(A-/HA)

2.45 = pKa + log(0.83375 /0.94875 )

pKa = 2.45 - log(0.83375 /0.94875 )

pKa = 2.506115

Ka = 10^-pKa = 10^-2.506115 = 0.00311 = 3.11*10^-3