How advanced is this class. If it is very beginning, I would answer,
.....H2SO4 ==> 2H^+ + SO4^2-
So 0.4M = 0.8M H^+.
If this is not a VERY VERY beginning class, that answer is not right. Why? Because the second ionization of H2SO4 is not 100% as is the first.
.....H2SO4 ==> H^+ +HSO4^-
This is 100% since H2SO4 is a strong acid. So (H+) from this is 0.4M.
There is a second ionization for HSO4^-. It is not strong.
........HSO4^- ==> H^+ + SO4^2-
I.......0.4........0......0
C.......-x.........x......x
E.......0.4-x......x......x
k2 = 0.012 = (H^+)(SO4^2-)/(HSO4^-)
Therefore, the H^+ will be the total of 0.4M from the first ionization and whatever the second one is BUT the amount from the second one is too small to change the total significantly. I can show you how to solve the second one if you wish but not tonight.
What is the hydrogen ion concentration, in mol dm−3, when 2.00 g of sulfuric acid is dissolved in water to give 0.400 dm3 of solution? could you explain in a simple and clear way please because im not very good at chemistry. thanks in advance
2 answers
i have never used the ice table before so i think its basic class. so do i just add those 2 numbers together? Thanks DrBob