How do I find the pH the molecule will have after finding the mass needed for a certain molarity?

Question

For example, I have this homework question part b, that I'm unsure how to go about.

"You have H2SO3 (d=1.02g/mL) and two solid salts (NaHSO3 and Na2SO3). (a) Figure out how to make 150 mM solutions of each. Report the amount (grams or mL) of sulfite species. (b) What are the pH of these solutions (part a!)?"

After making ice tables, I get stuck.

Answer 1

I'm unsure how big of a 150 mM (0.150 M) sample of each solution you need...that is not given in the problem. Let's say you need 100 mL of each. That is 0.100 L.

So for EACH:

1. calculate the molar mass. sulfurous acid (the first) is 82.07 g/mol. (this is a reference number)
2. calculate the mass of sulfurous acid using the molar mass from part 1. and the molarity (150 mM) from the problem and assuming 100 mL.

82.07 g/mol * 0.150 mol/L * 0.100 L = 1.23 g in 100 mL of water will give a 150 mM solution

You can use the density of sulfurous acid to calculate the volume instead:

1.23 g / 1.02 g/mL = 1.21 mL in 100 mL of water will give a 150 mM solution

You can repeat this for the two solids, but omit the density step (they aren't liquids).

The pH calculation:

1. For sulfurous acid, there is a value called pKa. Sulfurous acid has a pKa of 1.89 that is also related to the equilibrium constant, K. We can use this K value to calculate the pH.

H2SO3 + H2O <-> HSO3- + H3O+ where the equilibrium constant is K = 1.54x10^-2 (this is a reference number you can get from your book or google)

1. K = [HSO3-][H3O+]/[H2SO3] We can ignore water because it is present in such large quantities.

1. 1.54x10^-2 = x^2/0.150-x Here is where the ICE table came in!!! we can ignore -x when it is TINY. Remember your molarity is 150 mM or 0.150 M
2. Solving for x gives us [H3O+] = 4.8x10^-2

1. You can find pH = -log[H3O+] = 1.32

The problem is similar for each compound but K is different for different compounds and the molarity changes. Happy to clarify more!