[Sis290025]

1) The pH of a 0.009300 molar solution of unknown monoprotic acid was measured and found to be 2.990. Calculate the percentage dissociation of this acid.

% = [HA] dissociated /[HA] initial * 100 = 0.001023293/0.009300 M = 0.110031505 * 100 = 11.00%


[HA] dissociated = [H3O+] = antilog(-2.990) = 0.001023293

2) Vitamin C (ascorbic acid, C6H8O6, Ka = 8.0 x 10-5) is an organic acid that is essential to human health. You can buy vitamin C tablets in the grocery store in a range of different sizes. If a tablet containing 499.00mg of ascorbic acid were dissolved in enough water to prepare 269.00mL of solution, what would be the pH of the solution?

499.00 mg(1 g/1000 mg) = 0.499 g(1 mol/176.12 g) = 0.0028332955 mol vit. C

0.0028332955 mol vit. C/0.269 L = 0.10532697 M vit C

C6H8O6 + H2O <-> H3O+ + C6H7O6-

K_a = x^2/[0.10532697 - x] = 8.0*10^-5

x = 0.0029027847 = [H3O+] assuming 0.10532697 - x = 0.10532697

pH = -log(0.0029027847) = 2.5372 = 2.54

Thanks.

[Borek]

0.0028332955 mol vit. C/0.269 L = 0.10532697 M vit C

Check your math.

[Sis290025]

k_A = x^2/(0.010532697 - x)

x = sqrt(0.010532697*8.0*10^-5) = 9.1794104E-4 with same assumption

pH = -log(9.1794104E-4) = 3.0371852 = 3.04

Correct now?

[Borek]

Correct now?

Yes