What is the pH of a 50.0 mL solution containing 1.53 g of sodium sulfate. This is a diprotic base.

Answer 1

(Na2SO4) = grams/molar mass = approx 0.01 but that's just an estimate and that divided by 0.050 L = about 0.2 M

There are two ways to do this, the short way and the long way. The short way is not very exact but the long way is. The short way is to call Na2SO4 a salt of a strong base (NaOH) and a strong acid (H2SO4). Assuming that ALL of the H2SO4 ionizes completely into H^+ and SO4^2- ions then the pH is 7 for a neutral solution. since neither Na^+ nor SO4^2- is hydrolyzed.

However, H2SO4 actually has a K2 = about 0.012 and if you take that into account you get this from the hydrolysis of the sulfate.
..............SO4^2- + HOH ==> HSO4^- + OH^-
I..............0.2M...........................0.............0
C............-x.................................x.............x
E.........0.2-x................................x.............x

Kb for SO4^2- = (Kw/k2 for HSO4^-) = (x)(x)/(0.2-x)
Solve for x = (OH^-) and convert to pH. Remember that 0.2 is an estimate. This gives a pH of approx 7.7 and it requires that you solve the quadratic to get it; however, it is a little better answer than the 7 we assumed by the first method. If this is a beginning class I assume the prof wants the first solution. It it is an advanced class I think the prof will want the latter solution. kw is 0.012 in my text.

Answer 2

i dont understand where to start though

Answer 3

When I have shown you how to do this two ways, I don't understand what you don't understand. If you want to work it the short way you make the assumption I discussed and since neither Na^+ nor SO4^2- are hydrolyzed, then the pH is that of pure water which is 7.0.

If you think the prof wants you to work the problem the long way, then you start with that hydrolysis equation for SO4^2-, follow the ICE chart, and solve the equation I wrote for you. Substitute as shown and solve for OH^- and convert to pH.