Because both hydogens in H2SO4 are not strong acids. The first H ionizes completely (100%) so that 0.1M H2SO4 gives 0.1M H^+ and 0.1M HSO4^-. BUT the second H ion is not a strong acid (although it is relatively strong as weak acids go) and
HSO4^- ==> H^+ + SO4^2- with k2 = about 0.012.
So what happens is this.
BaSO4(s) ==> Ba^2+ + SO4^2-
If HCl is present, the H^+ from the strong (100% ionized) HCl, combines with the very small amount of SO4^2- to form the HSO4^- with the resulting k2. Now what happens with Le Chatelier's Principle? If H^+ is increased, HSO4^- forms, and the first reaction I showed moves to the right which increases the solubility of BaSO4. So if H^+ is too high, the solubility of BaSO4 is increased to the point that too much is lost due to solubility and the results for Ba^2+ or SO4^2- (depending upon which is being determined) will be too low. I have seen some calculations (and done some myself) and it is easy to lose as much as 1 mg BaSO4 and in many cases that is too much.
Show, with equations, why the solution from which BaSO4 is precipitated cannot be allowed to be more than 0.01-0.05M in HCl.
2 answers
I would love to see the equation of BaSO4 loss due to the increased HCl! 😁
0 answers