Question

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

Answer 1

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...

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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)`