# How to calculate the pH of a solution that is 0.02500M CO2..?

## CO2 (aq) + H2O ------> H2CO3 Khyd = 2.8 x 10^-3 H2CO3 + H2O ------> H3O+ + HCO3- K1 = 1.5 x 10^-4 HCO3- + H2O ------> H3O+ + CO32- K2 = 4.69 x 10^-11

Aug 20, 2017

You can do it like this:

#### Explanation:

Because the third ionisation is so small we shall only consider the first two:

$\textsf{C {O}_{2 \left(g\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{2} C {O}_{3 \left(a q\right)} \text{ } \textcolor{red}{\left(1\right)}}$

$\textsf{{H}_{2} C {O}_{3 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + H C {O}_{3 \left(a q\right)}^{-} \text{ } \textcolor{red}{\left(2\right)}}$

We know that:

$\textsf{{K}_{\text{hyd}} = \frac{\left[{H}_{2} C {O}_{3}\right]}{\left[C {O}_{2}\right]} = 2.8 \times {10}^{- 3}}$

$\textsf{{K}_{1} = \frac{\left[{H}_{3} {O}^{+}\right] \left[H C {O}_{3}^{-}\right]}{\left[{H}_{2} C {O}_{3}\right]} = 1.5 \times {10}^{- 4}}$

When you get successive equilibria like this you can add $\textsf{\textcolor{red}{\left(1\right)}}$ to $\textsf{\textcolor{red}{\left(2\right)} \Rightarrow}$

$\textsf{C {O}_{2 \left(g\right)} + {H}_{2} {O}_{\left(l\right)} + \cancel{{H}_{2} C {O}_{3 \left(a q\right)}} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s \cancel{{H}_{2} C {O}_{3 \left(a q\right)}} + {H}_{3} {O}_{\left(a q\right)}^{+} + H C {O}_{3 \left(a q\right)}^{-}}$

This becomes:

$\textsf{C {O}_{2 \left(g\right)} + 2 {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + H C {O}_{3 \left(a q\right)}^{-}}$

For which:

sf(K=([H_3O^+][HCO_3^-])/([CO_2])

It can be seen also that $\textsf{K = {K}_{\text{hyd}} \times {K}_{1}}$

$\therefore$$\textsf{K = 2.8 \times {10}^{- 3} \times 1.5 \times {10}^{- 4} = 4.2 \times {10}^{- 7}}$

Now we can set up an ICE table based on $\textsf{\text{mol/l"" } \Rightarrow}$

$\textsf{\text{ } C {O}_{2} + 2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H C {O}_{3}^{-}}$

$\textsf{I \text{ "0.02500" " " " " "0" " " } 0}$

$\textsf{C \text{ "-x" " " "+x" " " } + x}$

$\textsf{E \text{ "(0.02500-x)" " " "x" " " } x}$

$\therefore$$\textsf{\frac{{x}^{2}}{\left(0.02500 - x\right)} = K = 4.2 \times {10}^{- 7}}$

Because the ratio of $\textsf{\frac{\left[C {O}_{2}\right]}{K} > 500}$ we can assume that x is sufficiently small compared with 0.002500 such that $\textsf{\left(0.02500 - x\right) \Rightarrow 0.02500}$

$\therefore$$\textsf{{x}^{2} = 4.2 \times {10}^{- 7} \times 0.02500 = 1.05 \times {10}^{- 8}}$

$\textsf{x = \sqrt{1.05 \times {10}^{- 8}} = 1.024 \times {10}^{- 4}}$

$\therefore$$\textsf{\left[{H}_{3} {O}^{+}\right] = 1.024 \times {10}^{- 4} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}_{3} {O}^{+}\right] = - \log \left(1.024 \times {10}^{- 4}\right) = \textcolor{red}{3.99}}$