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

Question

Naomi Aronson · Accepted Answer

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: This is not an estimate. 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.) It should be noted that both hydrochloric acid and sulfuric acid are strong acids, with a 100% dissociation of their first proton in theory. 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"^(+))# But that's a different story about the second proton of sulfuric acid. 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)#,  an important 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)#

Noah Aldrich · Answer

undefined To find the pH of the solution, first calculate the moles of H+ ions from each acid. Then, determine the total moles of H+ ions in the solution. Finally, use the formula pH = -log[H+]. Given the concentrations and volumes, calculate the total moles of H+ ions and use it to find the pH.