Question

# A 0.872 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water...

A 0.872 gram sample of an unknown monoprotic acid is dissolved in 50.0 mL of water and titrated with a a 0.424 M aqueous sodium hydroxide solution. It is observed that after 9.40 milliliters of sodium hydroxide have been added, the pH is 3.350 and that an additional 5.50 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 Ka for the acid?

First, identify the molar mass

Total V for titration = V1+V2 = 9.4 + 5.5 = 14.9 mL

mmol of NaOH = MV = 0.424*14.9 = 6.3176 mmol of base

so, ther emust be also 6.3176 mmol of acid

MW = mas s/ mol = (0.872 g ) / (6.3176*10^-3 mol) = 138.027 g/mol

for Ka:

use:

pH = pKa + log(A-HA)

mmol of HA initially = 6.3176

mmol of NaOh added = MV = 9.4*0.424 = 3.9856

mmol of HA left = 6.3176-3.9856 = 2.332

mmol of A- formed = 0+3.9856

now

pH = pKa + log(A-/HA)

3.35 = pKa + log(3.9856/2.332)

pKa = 3.35 -  log(3.9856/2.332)

pKa = 3.117

Ka = 10^-´Ka = 10^-3.117

Ka= 7.638310^-4

#### Earn Coins

Coins can be redeemed for fabulous gifts.