#60 ml# #0.2(N) H_2SO_4+40 ml# #0.4(N) HCl# is given. What is the pH of the solution?

I got #"pH" = 0.60#. Notice how you can't really ignore the fact that the second proton on #"H"_2"SO"_4# will dissociate off of a weak acid.

If you assume that sulfuric acid dissociates 100% twice, you would get a total of #"0.006 mols" xx 2 = "0.012 mols H"^(+)# from #"H"_2"SO"_4#, which would give you a total #"H"^(+)# concentration of

#["H"^(+)] = "0.012 + 0.016 mols H"^(+)/("60 + 40 mL") xx "1000 mL"/"1 L"#

#~~# #"0.28 M"#

and consequently, an estimated #"pH"# of

#"pH" ~~ -log("0.28 M") = 0.55#,

which is about #7.7%# error. If your professor is OK with less than #5%# accuracy error, then #0.55# is not quite good enough...

DISCLAIMER: No approximations are made here.

Normality is defined with respect to the #"H"^(+)# or #"OH"^(-)# the solute dissociates into the solution.

So, a #"0.2 N"# solution of #"H"_2"SO"_4# is #"0.1 M"# with respect to #"HSO"_4^(-)#, for instance, because #"H"^(+)# is #2:1# with #"H"_2"SO"_4#. (The normality definition assumes 100% dissociation of all protons.)

Note that both sulfuric acid and hydrochloric acid are strong acids, which in principle have a 100% dissociation of their first proton.

If we calculate the mols of #H^(+)#, we can divide by the total volume last to find the final concentration.

#"H"_2"SO"_4(aq) -> "HSO"_4^(-)(aq) + "H"^(+)(aq)#

#"mols HSO"_4^(-) = "0.1 M" xx "0.060 L" = color(green)("0.006 mols first H"^(+))#

#"HCl"(aq) -> "H"^(+)(aq) + "Cl"^(-)(aq)#

#"mols H"^(+) = "0.4 M" xx "0.040 L" = color(green)("0.016 mols H"^(+))#

However, the second sulfuric acid proton is another story.

If we keep going without any approximations, we see that sulfuric acid dissociates into the weak acid, #"HSO"_4^(-)#, and that would have the following equilibrium:

#"HSO"_4^(-)(aq) + "H"_2"O"(l) rightleftharpoons "SO"_4^(2-)(aq) + "H"_3"O"^(+)(aq)#

#"I"" ""0.006 mols"" "-" "" "" "" ""0 mols"" "" ""0.006 mols"#
#"C"" "-x" "" "" "-" "" "" "+x" "" "" "" "+x#
#"E"" "0.006 - x" "-" "" "" "" "x" "" "" "" "0.006 + x#

#K_(a2) = 1.2 xx 10^(-2) = ((x)(0.006 + x))/(0.006 - x)#,

a non-negligible dissociation constant.

Solving this via the quadratic formula gives a physically reasonable #x# as

#x = "0.00337 mols second H"^(+)#

This means from #"H"_2"SO"_4#, you place

#"0.006 mols first H"^(+) + "0.00337 mols second H"^(+)#

#= "0.009 mols H"^(+)# into solution (to three decimal places).

The total mols of #"H"^(+)# in solution is then:

#overbrace("0.00937 mols H"^(+))^("H"_2"SO"_4) + overbrace("0.016 mols H"^(+))^("HCl") = "0.025 mols H"^(+)#

And by using the total volume of the solution, we then get the new concentration of #"H"^(+)# after the solution was mixed and prepared:

#"0.02537 mols H"^(+)/("60 + 40 mL") xx "1000 mL"/"1 L"#

#=# #"0.2537 M H"^(+)# at equilibrium

Therefore, the #"pH"# is:

#color(blue)("pH") = -log["H"^(+)]#

#= -log("0.2537 M")#

#= color(blue)(0.60)#