Question

Calculate the pH of a solution prepared by mixing 300mL of 0.25M sodium hydrogen ascorbate and 150mL of 0.2M HCl plus water to a final volume of 1 liter. The Ka of ascorbic acid is 5.0 x 10^-5

Answer 1

Given,

Concentration of C₆H₇NaO₆ = 0.25 M

Volume of C₆H₇NaO₆ = 300 mL x ( 1L /1000 mL) = 0.3 L

Concentration of HCl = 0.2 M

Volume of HCl = 150 mL x ( 1L /1000 mL) = 0.150 L

K_a of ascorbic acid = 5.0 x 10^-5

Calculating the number of moles of sodium hydrogen ascorbate and moles of HCl,

= 0.3 L x 0.25 M

= 0.075 mol of C₆H₇NaO₆

Similarly,

= 0.15 L x 0.2 M

= 0.03 mol of HCl

Now, the reaction between C₆H₇NaO₆ and HCl is,

C₆H₇NaO₆(aq) + HCl(aq)   C₆H₈O₆ + NaCl(aq)

Drawing an ICE chart,

	C6H7NaO6(aq)	HCl(aq)	C6H8O6
I(moles)	0.075	0.03	0
C(moles)	-0.03	-0.03	+0.03
E(moles)	0.045	0	0.03

Now, the new concentrations of C₆H₇NaO₆ and C₆H₈O₆ are,

[C₆H₇NaO₆] = 0.045 mol / 1 L = 0.045 M

[C₆H₈O₆] = 0.03 mol / 1 L = 0.03 M

Now, the equilbrium reaction for ascorbic acid is,

C₆H₈O₆(aq) + H₂O(l) C₆H₇O₆^-(aq) + H₃O⁺(aq)

Now, drawing an ICE chart,

	C6H8O6(aq)	C6H7O6-(aq)	H3O+(aq)
I(M)	0.03	0.045	0
C(M)	-x	+x	+x
E(M)	0.03-x	0.045+x	x

Now, K_a expression is,

K_a = [C₆H₇O₆^-] [H₃O⁺] / [C₆H₈O₆]

5.0 x 10^-5 = [0.045+x] [x] / [0.03-x]

5.0 x 10^-5 = [0.045] [x] / [0.03]--------- Here, [0.045+x]0.045 and [0.03+x]0.03, since x <<0.045,0.03

x = 3.33 x 10^-5

Now, from the ICe chart,

{H₃O⁺] =x = 3.33 x 10^-5

Now, we know, the formula to calculate the pH,

pH = -log[H₃O⁺]

pH = -log[ 3.33 x 10^-5]

pH = 4.48 Or 4.5 [ 2S.F]