Question

A 0.552-g sample of ascorbic acid was dissolved in water to a total volume of 0.20 mL and titrated with 0.1103 M KOH. The equivalence point occurred at 28.42 mL. The pH of the solution at 10.0mL of added base was 3.72. From this data, determine the molar rmass and Ka for vitamin C.

Answer 1

Answer:

176.1 g/mol , $Ka = 10^{-3.984} =1.03 * 10^{-4}$

Explanation:

The equation for the reaction is as shown in the attached diagram below

the reaction of Sodium Hydroxide, NaOH with ascorbic acid, C₆H₈O₆ is similar to the reaction of Potassium Hydroxide, KOH with ascorbic acid as show

C₆H₈O₆  +  KOH     ⇒     C₆H₇O₅K    +   H₂O

to produce one mole of potassium ascorbate and water

now,

0.1103 M KOH is contained in 1000mL

x moles is contained in 20mL

cross multiply making x the subject

No of moles of KOH = (0.1103 x 28.42)/1000 = 0.003135 moles

or

Moles KOH = 0.1103 x 0.02842 L = 0.003135  = moles ascorbic acid

Molar mass = 0.552 g / 0.003135 mol = 176.1 g/mol

Moles ascorbic acid = 0.552 / 176.1 =0.00313

moles NaOH = 0.0100 L x 0.1103 =0.001103

C₆H₈O₆ + OH⁻ >> C₆H₇O₆⁻ + H2O

Moles ascorbic acid in excess = 0.00313 - 0.001103 = 0.002027

Moles C₆H₇O₆⁻ = 0.001103

total volume = 20 + 10 = 30 mL = 0.030 L

concentration ascorbic acid = 0.002027 / 0.030 =0.0676 M

concentration C₆H₇O₆⁻ = 0.001103 / 0.030 =0.0368 M

pH = pKa + log [C₆H₇O₆⁻] / [C₆H₈O₆]

3.72 = pKa + log 0.0368 / 0.0676

3.72 = pKa - 0.264

pKa =3.984

$Ka = 10^{-3.984} =1.03 * 10^{-4}$

Ka = 10^-3.984 =1.03 x 10^-4