Question

The total of all carbonate species in lake water is 0.001 M, and the lake pH is 8.

Calculate the concentration of each of the individual carbonate species.

Answer 1

pH=-log[H3O+]

[H3O+]=10^-pH=10^-8 M

HCO3- + H2OCO32- + H3O+ pka2=10.3

H2CO3+ H2OHCO3- + H3O+ pka1=6.35

as pka2>pH so no deprotonation of HCO3- took pace. Species in lake water, H2CO3 , HCO3-

pH=pka1+log [base]/[acid]

8=6.35 +log [HCO3-]/[H2CO3]

8=6.35+log [HCO3-]/[H2CO3]                  

1.65=log [HCO3-]/[/[H2CO3]

10^1.65=[HCO3-]/[/[H2CO3]=44.67

[HCO3-]/[H2CO3]=44.67

As total conc =0.001M

[HCO3-]/ [H2CO3] = 44.67

[HCO3-] + [H2CO3] = 0.001M

44.67[H2CO3]+ [H2CO3]= 0.001M

45.67[H2CO3]=0.001M

[H2CO3]==0.001/45.67=2.18*10^-5 M

[HCO3-] = 44.67[H2CO3]                                  

[HCO3-]=44.67(2.18*10^-5 M)=9.74*10-4M

[H2CO3]= 2.18*10^-5 M

[HCO3-]=9.74*10-4M