Question

Carbonic acid is a diprotic acid, H2CO3, with Ka1 = 4.2 × 10 –7 and Ka2 = 4.8 × 10 –11 at 25°C. The ion product for water is Kw = 1.0 × 10 –14 at 25°C. What is the OH– concentration of a solution that is 0.37 M in Na2CO3?

A. 4.2 × 10–6 M
B. 2.1 × 10–4 M
C. 8.8 × 10–3 M
D. 9.4 × 10–5 M
E. 3.9 × 10–4 M

Answer 1

Without additional information, we cannot accurately calculate the exact OH− concentration of the 0.37 M Na2CO3 solution since we only have the Na2CO3 concentration and the second ionization constant for carbonic acid.

Step-by-step explanation:

To calculate the OH− concentration of a solution that is 0.37 M in Na2CO3, we need to consider the ionization of carbonate that comes from Na2CO3 in water. Na2CO3 completely dissociates into 2 Na+ and CO32−. The carbonate ion will then further react with water to form bicarbonate (HCO3−) and hydroxide ions (OH−), as follows:

HCO3− (aq) + H2O(l) ⇌ OH− (aq) + H2CO3 (aq)

To find the hydroxide ion concentration, we must consider the second ionization constant Ka2 = 4.8 × 10−11 and the fact that [CO32−] from the dissociation of Na2CO3 is initially 0.37 M. Since Ka2 is very small, the change in concentration due to the second ionization is negligible compared to the initial concentration. Therefore, we can approximate that [OH−] is roughly equal to the initial concentration of CO32−, which is 0.37 M.

Finally, we use the ion product for water (Kw = 1.0 × 10−14) to find the concentration of OH−: [OH−] = Kw / [H+]. However, since we only have the concentration of Na2CO3, we cannot directly solve for [OH−] without additional information. Therefore, from the provided information and without doing further calculations, we are unable to determine the exact concentration of OH− in the solution.