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

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))#