A 8.80-g sample of homogentisic acid, a weak organic acid having ka=4.0 x 10^-5,

Question

A 8.80-g sample of homogentisic acid, a weak organic acid having ka=4.0 x 10^-5, is dissolved in 25.00 ml of water and its pH is measured to be 2.039. What is the molar mass of homogentisic acid?
A) 352 g/mol B) 4.20 g/mol C) 168 g/mol D) 2.09 g/mol E) 964 g/mol A 8.80-g sample of homogentisic acid, a weak organic acid having ka=4.0 x 10^-5, is dissolved in 25.00 ml of water and its pH is measured to be 2.039. What is the molar mass of homogentisic acid?
A) 352 g/mol B) 4.20 g/mol C) 168 g/mol D) 2.09 g/mol E) 964 g/mol
A) 352 g/mol B) 4.20 g/mol C) 168 g/mol D) 2.09 g/mol E) 964 g/mol

Explanation / Answer

The weak acid will dissociate as

                                        HA -->           H+        +          A-

Inital concentration             a                  0                         0

Change in conc                  -x               x                          x

Equilbirium conc              a-x               x                           x

Given

pH = 2.039 = -log[H+]

so [H+] = 0.00914 M

So [H+] = x = 0.00914

Ka = [H+] [A-] / [HA]

Ka = 4.0 x 10^-5 = 0.00914 X 0.00914 / a-0.00914

we can ignore x with respect to a

4.0 x 10^-5 = 0.00914 x 0.00914 / a

Concentration of acid = a = 2.088 M

we know that

concentration = mass of solute X 1000 / mol wt of solute x volume of solution

2.088 = 8.8 X 1000 / Mol wt X 25

mol wt = 168.58 or approx 168 grams / mole

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