Question

a) The consecutive equilibria:

CO2(g) <=>CO2(aq) + H2O(l) <=>H2CO3(aq) <=>H+(aq) + HCO3-(aq), explain why:

A. carbonated beverages are less acidic than pure water.

B. carbonated beverages contain measurably less water than the same volume of pure water.

C. rainwater is slightly acidic

b) A 1.0 M solution of a weak monoprotic acid with Ka = 1.0 10-6 would have a pH of:

A. 1 B. 2 C. 3 D. 6

c) Ascorbic acid is a diprotic acid with Ka1 = 1.0 10-5, and Ka2 = 5 10-12. The pH of a 0.1 M solution of ascorbic acid is closest to:

A. 7.0 B. 3.0 C. 10-5 D. 2.8

Answer 1

a) A) The pKa of HCO3- is greater than 7, therefore maintaining the equilibrium at basic pH. HCO3- is dissolved in carbonated drinks.

B) Carbonated drinks contain gases and other substances (CO2, HCO3-, etc.) dissolved, which means that not all volume is pure water.

C) Rainwater contains traces of H2S, which is an acid compound and gives that property to rainwater.

b) From the expression of Ka we have:

Ka = [H3O +] * [A-] / [HA] = X ^ 2/1 - X

It is assumed that - X is negligible and clears:

X = [H3O +] = √Ka * 1 = √1x10 ^ -6 * 1 = 1x10 ^ -3 M

It is calculated pH = - log [H3O +] = - log 1x10 ^ -3 = 3

Option C.

c) Same previous procedure we have:

X = [H3O +] = √Ka1 * 0.1 M = √1x10 ^ -5 * 0.1 = 1x10 ^ -3 M

pH = - log 1x10 ^ -3 = 3

Option B.

If you liked the answer, please rate it in a positive way, you would help me a lot, thank you.