# Dilution of a diprotic acid?

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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?

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?

##### 1 Answer

I think you were right, if you read the question correctly. I got

Well, let's see.

#"10 cm"^3 = "10 mL"#

#"100 cm"^3 = "100 mL"#

The starting

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

#"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

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

And this gives a

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

So this is fine so far. The

By diluting it by a factor of

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

But by doing this we ask the equilibrium to shift to lower

#"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

This means that

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

And the new

#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