[hhzhang]

I encountered a question:
Calculate the solubility of BaCO₃ in a solution of pH 7.00.
K_sp for BaCO₃ is 5.0×10^-9.
For H₂CO₃, K_a1 = 4.45×10^-7 and K_a2 = 4.69×10^-11.

And this is the given answer:
pH = 7.00, therefore [H⁺] = 1.0×10^-7 M

α₂ = 3.8 ×10^-4

Solubility,
$$s = \sqrt{\frac{K_{sp}}{\alpha_2}} = 3.6 \times 10^{-3}$$

Solubility of BaCO₃ in pH 7.00 = 3.6×10^-3 M.
Solubility of BaCO₃ in water = 7.1×10^-5 M.

My question is, pH of water is also pH 7.00, but why we don't consider the hydrolysis of CO₃^2- here?

**Sorry but I don't know how to type the equation, may someone helps? ><"**

[Borek]

You forgot to add [tеx][/tеx] tags.

K_sp is K_{sp}, not K_s_p (the latter seems to be ambiguous).

To the problem: what is α₂?

[hhzhang]

Thank you very much!!! I spent an hour to deal with that equation but gave up ><"

$$\alpha_2 = \frac{K_{a1}K_{a2}}{[H^{+}]^{2} + K_{a1}[H^{+}] + K_{a1}K_{a2}}$$

$$= 3.8 \times 10^{-4}$$

Here it is!

[Borek]

And what is the meaning of α₂?

[hhzhang]

For diprotic acid H₂CO₃,

Total concentration of diprotic acid species, C_H2CO3 = [H₂CO₃] + [HCO₃^-] + [CO₃^2-]

and the fraction of diprotic acid species, $$\alpha_2$$ is given by the formula:

$$\alpha_2 = \frac{[CO_{3}^{2-}]}{C_{H_2CO_3}}$$

$$[CO_{3}^{2-}] = \alpha_2 C_{H_2CO_3}$$

Replacing it into solubility product of BaCO₃,

K_sp = [Ba²⁺][CO₃^2-]

= [Ba²⁺]α₂ C_H2CO3

[Borek]

fraction of diprotic acid species, $$\alpha_2$$ is given by the formula:

$$\alpha_2 = \frac{[CO_{3}^{2-}]}{C_{H_2CO_3}}$$

It tells what fraction of the acid is in the form of CO₃^2- as a function of total concentration and pH, after system reached equilibrium - so all reactions like hydrolysis are already taken into account.