What is the OH^- concentraton of a solution that is 0.10M in Na2C2O4?

I have a couple of these concentration questions asking different things but I don't know how to do them

For salts, like this one, you hydrolyze them; i.e., react them with water. Suppose we have 0.1 M NaA (the salt of NaOH and a weak acid). The general equation is

A^- + HOH ==> HA + OH^-

Kb = Kw/Ka = (HA)(A^-)/(HA)

You know Kw and Ka, (A^-) = 0.1 M and (HA)= (OH^-) = x
You solve for x, which in this case is (OH^-), take -log to obtain pOH, then subract from 14 to get pH.

To find the OH^- concentration of a solution that is 0.10M in Na2C2O4, you need to determine the moles of OH^- ions produced from the dissociation of Na2C2O4.

The balanced chemical equation for the dissociation of Na2C2O4 is:
Na2C2O4 → 2Na+ + C2O4^2-

Since each C2O4^2- ion has 2 negative charges, it can combine with 2H+ ions from water to form H2C2O4 (oxalic acid) and 2OH^- ions. Therefore, the concentration of OH^- ions is twice the concentration of Na2C2O4.

Given that the concentration of Na2C2O4 is 0.10M, the concentration of OH^- ions would be:

OH^- concentration = 2 × 0.10M
OH^- concentration = 0.20M

Therefore, the OH^- concentration of the solution is 0.20M.

To determine the OH^- concentration of the solution, you need to understand the dissociation of Na2C2O4 in water and the relationship between the concentration of hydroxide ions (OH^-) and the concentration of the compound.

The compound Na2C2O4, also known as sodium oxalate, dissociates in water according to the following balanced chemical equation:

Na2C2O4(l) → 2 Na+(aq) + C2O4^2-(aq)

From this equation, we can see that for every one mole of Na2C2O4 dissolved, it produces two moles of Na+ ions and one mole of C2O4^2- ions.

However, the concentration of hydroxide ions (OH^-) is not directly related to the concentration of Na2C2O4. To calculate the OH^- concentration, we need to consider the hydrolysis of the C2O4^2- ions.

The C2O4^2- ion reacts with water to form OH^- ions and a weak acid called oxalic acid (H2C2O4) according to the reaction:

C2O4^2-(aq) + H2O(l) ⇌ 2 OH^-(aq) + H2C2O4(aq)

The OH^- ions are produced in a 1:2 ratio with respect to the C2O4^2- ions, meaning that for every one mole of C2O4^2- ions that hydrolyzes, two moles of OH^- ions are produced.

Now, to find the OH^- concentration, we need to know the initial concentration of Na2C2O4. Since the compound dissolves completely in water and dissociates into its ions, the initial concentration of C2O4^2- ions is also 0.10 M.

Thus, from the hydrolysis reaction, we know that for every mole of C2O4^2-, two moles of OH^- ions are produced. Therefore, the concentration of OH^- ions in this solution is twice the concentration of C2O4^2- ions.

OH^- concentration = 2 * C2O4^2- concentration

OH^- concentration = 2 * 0.10 M

OH^- concentration = 0.20 M

Therefore, the OH^- concentration of the solution is 0.20 M.