[Rutherford]

In a saturated solution (V=100 ml) of an acid (Mr=212, Ka=1.5·10^-3), the concentration of HA and A^- dissolved is 0.0254 M and 0.000616 M respectively (the total concentration is 0.0254 M + 0.000616 M = 0.0316 M). How many grams of NaOH should be added to this solution so that the solubility increases twice, i.e. the total concentration becomes 2· 0.0316 M = 0.0632 M?

HA + NaOH NaA + H₂O
Some of the HA reacted (this amount I will mark with x), some dissociated (I mark this with y). A^- is obtained in both processes, H⁺ only in dissociation. Then writing the equilibrium expression:
1.5·10^-3=(x+y)·y/(0.0632-x-y)
I have two unknowns. I need to calculate x to get the mass of NaOH. How to solve this?

[Rutherford]

I got the correct answer m=0.147 g by assuming that 0.0632-x-y = 0.0254 M (which is the concentration of non-dissociated HA in the saturated solution without NaOH). Why is the concentration of non-dissociated HA equal in the saturated solution and when NaOH is added? Why isn't it lower?

EDIT: I think that it has to be assumed that the second solution will be saturated, too. That would do it.

[Borek]

Definitely in both solutions HA must be identical, that's how I read the question.

[Rutherford]

It isn't explicitly said that the second solution is saturated, too.

[Big-Daddy]

0.0254 M + 0.000616 M = 0.0316 M).

0.0254 + 0.000616 = 0.026016 M

I also got a different final answer. Can you link the problem so I can check to see if my working is consistent?

I thought [HA] would be the same in both solutions and so [A^-] = 0.026632 M in the new solution (since the total concentration of acid has doubled for its solubility to double). I thus got [H⁺] = 1.43 * 10^-3 M in the new solution. Then the initial concentration of NaOH must be 0.0252 M, which can be converted to mass.

[Rutherford]

Sorry, that was one extra zero. It should be 0.00616 M. The problem can't be found in English.

[Borek]

It isn't explicitly said that the second solution is saturated, too.

If it is not saturated question doesn't make sense.

[Rutherford]

Isn't it possible to make an unsaturated solution this way?

[Borek]

How are you going to calculate amount of NaOH then? Assuming solution is saturated you can calculate minimum amount of NaOH required. You can use excess NaOH and just enough acid to produce the required solution, but that's trivial - besides, even then, to know what it means "excess NaOH" you have to calculate minimum amount of NaOH - and as stated above, that requires assuming solution is saturated.

[Rutherford]

I know it's then impossible to calculate the answer, but I meant trivial. Okay. Thank you for the explanation.
Calculate minimum amount of NaOH wouldn't confuse me.