Question

A 0.1L 0.5M solution of sulphuric acid is mixed with 0.2L 0.1 M phosphoric acid , the concentration of H+ ions in the resultant solution?

Answer 1

Is this a low level or an advanced chemistry class? Assuming advanced, I would do this. H2SO4 is a strong acid for the first ionization but k2 for H2SO4 is 0.0102. H3PO4 has 3 k values but only the first one is large enough to matter. k1 is 0.00711. My rationale would be that k2 for H2SO4 is stronger than k1 for H3PO4 so calculate H^+ from H2SO4, use that to solve for H^+ for k1 for H3PO4, add mols ^+ from H2SO4 to mols H^+ from H3PO4, then final M H^+ = mols/L. Here is how you do H^+ from H2SO4.
......H2SO4 ==> H^+ + HSO4^- (STRONG)
I......0.5......0......0
C.....-0.5.....0.5....0.5
E.......0......0.5....0.5

Then HSO4^- ==> H^+ + SO4^2-
I....0.5.......0.5.....0
C.....-x........x......x
E...0.5-x....0.5+x.....x

k2 = (0.5+x)(x)/(0.5-x)
Solve for x and I get 0.0098 (use the quadratic formula for this) and this rounds to about 0.01. That added to 0.5 gives 0.51M for the H2SO4. Use this for H^+ for H3PO4 and the H^+ contribution from H3PO4(that's the same as H2PO4^-) is negligible.
Then millimols H^+ from H2SO4 = mL x M
and millimols H^+ from H3PO4 negligible.
Finally, in final solution M H^+ = mmols/mL. Post your work if you get stuck.