# 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?

Dec 21, 2017

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

Well, let's see.

$\text{10 cm"^3 = "10 mL}$
$\text{100 cm"^3 = "100 mL}$

The starting $\text{pH}$ was $2$, supposedly, so let's verify that.

This is due to the first dissociation:

${\text{H"_2"SO"_4(aq) " "" "->" "" " "HSO"_4^(-)(aq) + "H}}^{+} \left(a q\right)$

$\text{I"" ""0.005 M"" "" "" "" "" "" ""0 M"" "" "" "" ""0 M}$
$\text{C"" "- x_1" "" "" "" "" "" "" "+x_1" "" "" } + {x}_{1}$
$\text{E"" "(0.005 - x_1)" M"" "" "" "x_1" M"" "" "" "x_1" M}$

where ${x}_{1} \approx \text{0.005 M}$, and also the second dissociation:

${\text{HSO"_4^(-)(aq) " "" "rightleftharpoons" "" " "SO"_4^(2-)(aq) + "H}}^{+} \left(a q\right)$

$\text{I"" ""0.005 M"" "" "" "" "" "" ""0 M"" "" "" "" ""0.005 M}$
$\text{C"" "- x_2" "" "" "" "" "" "" "+x_2" "" "" } + {x}_{2}$
$\text{E"" "(0.005 - x_2)" M"" "" "" "x_2" M"" "" "0.005 + x_2" M}$

where ${K}_{a 2} = 1.2 \times {10}^{- 2}$.

From this,

$1.2 \times {10}^{- 2} = \frac{{x}_{2} \left(0.005 + {x}_{2}\right)}{0.005 - {x}_{2}}$

Solving this gives ${x}_{2} = \text{0.003 M}$. As a result,

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

And this gives a $\text{pH}$ of

$\text{pH} = - \log \left(0.008\right) = 2.10$

So this is fine so far. The $\text{pH}$ 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 ${\text{H"_2"SO}}_{4}$ is nearly all gone so that is not affected for our intents and purposes.

That means

${\left[{\text{HSO}}_{4}^{-}\right]}_{i 2} = \frac{0.005 - {x}_{2}}{10} = \frac{0.005 - 0.003}{10}$

$=$ $\text{0.0002 M}$

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

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

with $\text{pH} = 3.10$.

But by doing this we ask the equilibrium to shift to lower $\text{pH}$. This new shift is given as:

${\text{HSO"_4^(-)(aq) " "" "rightleftharpoons" "" " "SO"_4^(2-)(aq) + "H}}^{+} \left(a q\right)$

$\text{I"" ""0.0002 M"" "" "" "" "" ""0.0003 M"" "" ""0.0008 M}$
$\text{C"" "- x_3" "" "" "" "" "" "" "+x_3" "" "" } + {x}_{3}$
$\text{E"color(white)(.)(0.0002 - x_3)" M"color(white)(..)0.0003"+"x_3" M"" ""0.0008+"x_3" M}$

And so:

${K}_{a 2} = 1.2 \times {10}^{- 2} = \frac{\left(0.0003 + {x}_{3}\right) \left(0.0008 + {x}_{3}\right)}{0.0002 - {x}_{3}}$

from which we find via Wolfram Alpha has ${x}_{3} = \text{0.00016 M}$ (cannot use small x approximation).

This means that

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

And the new $\text{pH}$ is:

$\textcolor{b l u e}{\text{pH}} = - \log \left(0.00096\right) = \textcolor{b l u e}{3.02}$

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