Question

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

Answer 1

You can do it like this:

Explanation:

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

`sf(CO_(2(g))+H_2O_((l))rightleftharpoonsH_2CO_(3(aq))" "color(red)((1)))`

`sf(H_2CO_(3(aq))+H_2O_((l))rightleftharpoonsH_3O_((aq))^(+)+HCO_(3(aq))^(-)" "color(red)((2)))`

We know that:

`sf(K_("hyd")=([H_2CO_3])/([CO_2])=2.8xx10^(-3))`

`sf(K_(1)=([H_3O^+][HCO_3^-])/([H_2CO_3])=1.5xx10^(-4))`

When you get successive equilibria like this you can add `sf(color(red)((1)))` to `sf(color(red)((2))rArr)`

`sf(CO_(2(g))+H_2O_((l))+cancel(H_2CO_(3(aq)))+H_2O_((l))rightleftharpoonscancel(H_2CO_(3(aq)))+H_3O_((aq))^(+)+HCO_(3(aq))^(-))`

This becomes:

`sf(CO_(2(g))+2H_2O_((l))rightleftharpoonsH_3O_((aq))^(+)+HCO_(3(aq))^(-))`

For which:

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

It can be seen also that `sf(K=K_("hyd")xxK_(1))`

`:.` `sf(K=2.8xx10^(-3)xx1.5xx10^(-4)=4.2xx10^(-7))`

Now we can set up an ICE table based on `sf("mol/l"" "rArr)`

`sf(" "CO_2+2H_2OrightleftharpoonsH_3O^(+)+HCO_3^-)`

`sf(I" "0.02500" " " " " "0" " " "0)`

`sf(C" "-x" " " "+x" " " "+x)`

`sf(E" "(0.02500-x)" " " "x" " " "x)`

`:.` `sf((x^2)/((0.02500-x))=K=4.2xx10^(-7))`

Because the ratio of `sf(([CO_2])/(K)>500)` we can assume that x is sufficiently small compared with 0.002500 such that `sf((0.02500-x)rArr0.02500)`

`:.` `sf(x^2=4.2xx10^(-7)xx0.02500=1.05xx10^(-8))`

`sf(x=sqrt(1.05xx10^(-8))=1.024xx10^(-4))`

`:.` `sf([H_3O^+]=1.024xx10^(-4)color(white)(x)"mol/l")`

`sf(pH=-log[H_3O^+]=-log(1.024xx10^(-4))=color(red)(3.99))`