Determine mass of sodium carbonate decahydrate from the pH?

Question

Check my answer, also check sig figs.

Determine the mass of sodium carbonate decahydrate necessary to make 500.00 mL of a solution with a pH = 11.50.

first we change pH to [H+] 10^-11.50= 3.16x10^-12 this is the molarity to get the mols we multiply this by the volume in liters so,

3.16x10^-12 x .500L = 1.58x10^-12 mol

now to get the pass we take the mol and multiply by the molar mass. the formula is Na2CO3 * 10H20

Im not sure if you add the 10H20 but I did, can anyone tell me?

1.58x10^-12 * 285 = 4.52 x 10^-10g is the mass

Answer 1

Absolutely wrong. The Na2CO3 salt dissociates in water solutions:

Na2CO3 --> 2Na+ + (CO3)^(-2)

Then the CO3(2-) anion reacts with water:

CO3(2-) + H2O <==> HCO3(-) + OH-

(That's why the solution has pH > 7. Another equilibrium also takes place HCO3(-) + H2O <==> H2CO3 + OH- but you can ignore it because it has a very small Kb)

Now you have to know the Kb (or the pKb) constant of the above equilibrium. I found that it equals Kb = 2*10^(-4).

If C the unknown Molarity of the solution and [HCO3(-)] = [OH-] = x, then:

Kb = [HCO3-]*[OH-]/[CO3(2-], or Kb = x^2/(C - x), or approx.

Kb = x^2/C ==> C = x^2/Kb. But you can find x from the pH:

pH = 11.5, pOH = 14 - pH, pOH = 14 - 11.5, pOH = 2.5, hence

x = [OH-] = 10^(-2.5). So:

C = 10^(-5)/2*10^-4), C = 0.05 M.

Now find the moles of Na2CO3:

n = C*V, n = 0.05*0.5, n = 0.025 mol

The molar mass of Na2CO3.10H2O equals Mr = 286 g/mol, so the mass of the salt is:

m = n*Mr, n = 0.025*286, n = 7.15 g.

I think that's the right answer.

Answer 2

Sodium carbonate needs to be treated as a weak base. This is a very long answer: please bear with me and the inability to use some math symbols.

Equilibrium equation:
Na2CO3 + H2O <--> NaHCO3 + NaOH
CO3 + H2O <--> HCO3 + OH.

You can also write K = [OH][HCO3]/[CO3]. This K is the Kb. The pKa for this reaction is 10.33. Remember that Kb*Ka = 10^-14 and pKa + pKb = 14.
pKb = 14 - 10.33 = 3.67.
Kb = 10^-3.67.
You also know [OH]. [OH] = 10^-14 / [H]. We know [H] is 10^-11.50, so [OH] = 10^-2.500.
[CO3] is not known. Remember you must take into account that you have lost 10^-2.500 moles/L of CO3 that you put in. So [CO3] = x - 10^-2.500

Let's plug these values in:
10^-3.67 = (10^-2.500)^2 / (x - 10^-2.500).
10^-3.67 (x - 10^-2.500) = 10^-5.000. W
x - 10^-2.500 = 10^-1.33

At this point you should continue as you did before.
4.99*10^-2M * 0.5L = 2.50^-2 mol * 285 = 7.12g Na2CO3 * 10H2.