Question

Dilution of a diprotic acid?

10 cm^3 of a 5 x10^-3 mol dm^-3 sulfuric acid is added to an empty volumetric flask. It's pH is currently 2. The solution is the diluted so the total volume is made up to 100 cm^3 with distilled water, what is the new pH of the solution?
According to my book the answer should be 4, but I can seemingly only get it to 3. Help?

Answer 1

I think you were right, if you read the question correctly. I got `3.02` doing it the rigorous way with no big approximations.

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Well, let's see.

`"10 cm"^3 = "10 mL"`
`"100 cm"^3 = "100 mL"`

The starting `"pH"` was `2`, supposedly, so let's verify that.

This is due to the first dissociation:

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

`"I"" ""0.005 M"" "" "" "" "" "" ""0 M"" "" "" "" ""0 M"`
`"C"" "- x_1" "" "" "" "" "" "" "+x_1" "" "" "+x_1`
`"E"" "(0.005 - x_1)" M"" "" "" "x_1" M"" "" "" "x_1" M"`

where `x_1 ~~ "0.005 M"`, and also the second dissociation:

`"HSO"_4^(-)(aq) " "" "rightleftharpoons" "" " "SO"_4^(2-)(aq) + "H"^(+)(aq)`

`"I"" ""0.005 M"" "" "" "" "" "" ""0 M"" "" "" "" ""0.005 M"`
`"C"" "- x_2" "" "" "" "" "" "" "+x_2" "" "" "+x_2`
`"E"" "(0.005 - x_2)" M"" "" "" "x_2" M"" "" "0.005 + x_2" M"`

where `K_(a2) = 1.2 xx 10^(-2)`.

From this,

`1.2 xx 10^(-2) = (x_2(0.005 + x_2))/(0.005 - x_2)`

Solving this gives `x_2 = "0.003 M"`. As a result,

`["H"^(+)]_(eq) = "0.005 M" + "0.003 M" = "0.008 M"`

And this gives a `"pH"` of

`"pH" = -log(0.008) = 2.10`

So this is fine so far. The `"pH"` is indeed close to `2`.

By diluting it by a factor of `10`, however, the concentration of EVERYTHING in solution decreases by a factor of `10`. The `"H"_2"SO"_4` is nearly all gone so that is not affected for our intents and purposes.

That means

`["HSO"_4^(-)]_(i2) = (0.005 - x_2)/10 = (0.005 - 0.003)/10`

`=` `"0.0002 M"`

`["H"^(+)]_(i2) = ("0.005 + 0.003 M")/(10) = "0.0008 M"`

`["SO"_4^(2-)]_(i2) = "0.003 M"/10 = "0.0003 M"`

with `"pH" = 3.10`.

But by doing this we ask the equilibrium to shift to lower `"pH"`. This new shift is given as:

`"HSO"_4^(-)(aq) " "" "rightleftharpoons" "" " "SO"_4^(2-)(aq) + "H"^(+)(aq)`

`"I"" ""0.0002 M"" "" "" "" "" ""0.0003 M"" "" ""0.0008 M"`
`"C"" "- x_3" "" "" "" "" "" "" "+x_3" "" "" "+x_3`
`"E"color(white)(.)(0.0002 - x_3)" M"color(white)(..)0.0003"+"x_3" M"" ""0.0008+"x_3" M"`

And so:

`K_(a2) = 1.2 xx 10^(-2) = ((0.0003 + x_3)(0.0008 + x_3))/(0.0002 - x_3)`

from which we find via Wolfram Alpha has `x_3 = "0.00016 M"` (cannot use small x approximation).

This means that

`["H"^(+)]_(eq2) = 0.0008 + 0.00016 = "0.00096 M"`

And the new `"pH"` is:

`color(blue)("pH") = -log(0.00096) = color(blue)(3.02)`

So even doing it the rigorous way, it is as you expected. It should be `3`!