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?
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