Question

A 1.31 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water and titrated with a a 0.496 M aqueous sodium hydroxide solution. It is observed that after 9.16 milliliters of sodium hydroxide have been added, the pH is 7.227 and that an additional 14.6 mL of the sodium hydroxide solution is required to reach the equivalence point.

(1) What is the molecular weight of the acid? ____ g/mol

(2) What is the value of K_a for the acid?

Answer 1

NAOH --> 0.496 M

Vtotal = 9.16+14.6 = 23.76mL

total mol of NaOH required= MV = (0.496*23.76) = 11.78496 mmol of NaOH

if monoprotic, then

mmol of acid = mmol of base = 11.78496

mmol of acid = 11.78496 mmol

1)

molar mass = mass of acid / mol of acid = (1.31/(11.78496*10^-3)) = 111.158 g/mol

2)

for pKa

use buffer equation so

pH = pKa + log(A-/HA)

we need, A- and HA left when V = 9.16 mL was added

so

mmol of base added at 9.16 mL = MV = 0.496*9.16 = 4.54336 mmol of base

mmol of acid initially = MV = 1.31/111.158 *10^3 = 11.78 mmol

note that base reacts iwth acid to form:

conjugate base (A-) =4.54336 mmol formed

acid left = 11.78-4.54336 = 7.23664 mmol

substitute

pH = pKa + log(A-/HA)

7.227 = pKa + log(4.54336 /7.23664 )

pK = 7.227 - log(4.54336 /7.23664 )

pKa= 7.42915