5. A water sample with a total alkalinity of 2.00 x 10 moles/L has a pH of 7.00. Compute the equilibrium concentrations of CO2, HCOs , co,, and OH Note that water and carbon dioxide can replace carbonic acid in the Kai reaction without changing the value of Kal H2O + CO2 H+ + HCO, Ka1 = 4.5 x 10-7 Solution The equilibrim equations involved are: H2O +CO2 <---->H+ +HCO3- ka1=4.5*10^-7 HCO3- <--->CO32- +H+ ka2=4.8*10^-11 Also given, pH=-log [H+]=7.0 [H+]=10^-7 M ...........(a) [OH-]=kw/[H+]=10^-14/[H+]=10^-14.............(b) pH=pka +log [base]/[acid] [henderson-hasselbach equation] pka2=-log ka2=-log(4.8*10^-11)=10.3 7.0=10.3 +log[CO32-]/[HCO3-] or, [CO32-]/[HCO3-]=5.012e-4. or,[CO32-]=(5.012e-4.)[HCO3-]..................(1) Also, H2O +CO2 <---->H+ +HCO3- ka1=4.5*10^-7=[H+][HCO3-]/[CO2] for x=[H+]=[HCO3-] (at equilibrium) [CO2]=[CO2]o-x , [CO2]o=initial concentration,but x is very less for very less dissociation ( ka1 is very small) [CO2]eq=[CO2]o 4.5*10^-7=[H+][HCO3-]/[CO2]=x^2/[CO2] or,[CO2]eq=x^2/(4.5*10^-7)=[HCO3-]eq/(4.5*10^-7) [CO2]eq=[HCO3-]eq^2/(4.5*10^-7) ...................(2) total alkalnity=[HCO3-] +2[CO32-] +[OH-] -[H+]=2.0*10^-3 mol/L plugging in eqn (a),(b),(1) & (2) in the equation for total alkalinity, total alkalnity=[HCO3-] +2(5.012e-4.)[HCO3-]. =2.0*10^-3 mol/L or, [HCO3-]=1.997*10^-3 mol/L [CO32-]eq=(5.012e-4.)[HCO3-].=(5.012e-4.)(2.0*10^-3 mol/L)=1.002*10^-6 mol/l [H+]eq=[OH-]eq=10^-7 mol/L [CO2]=[HCO3-]eq^2/(4.5*10^-7) =(1.997*10^-3 mol/L)^2/(4.5*10^-7)=8.862 mol/L [CO2]eq=8.862*10^3 mol/L.

1. 5. A water sample with a total alkalinity of 2.00 x 10 moles/L has a pH of 7.00. Compute the
equilibrium concentrations of CO2, HCOs , co,, and OH Note that water and carbon dioxide can
replace carbonic acid in the Kai reaction without changing the value of Kal H2O + CO2 H+ +
HCO, Ka1 = 4.5 x 10-7
Solution
The equilibrim equations involved are:
H2O +CO2 <---->H+ +HCO3- ka1=4.5*10^-7
HCO3- <--->CO32- +H+ ka2=4.8*10^-11
Also given, pH=-log [H+]=7.0
[H+]=10^-7 M ...........(a)
[OH-]=kw/[H+]=10^-14/[H+]=10^-14.............(b)
pH=pka +log [base]/[acid] [henderson-hasselbach equation]
pka2=-log ka2=-log(4.8*10^-11)=10.3
7.0=10.3 +log[CO32-]/[HCO3-]
or, [CO32-]/[HCO3-]=5.012e-4.
or,[CO32-]=(5.012e-4.)[HCO3-]..................(1)
Also, H2O +CO2 <---->H+ +HCO3- ka1=4.5*10^-7=[H+][HCO3-]/[CO2]
for x=[H+]=[HCO3-] (at equilibrium)
[CO2]=[CO2]o-x , [CO2]o=initial concentration,but x is very less for very less dissociation ( ka1
is very small)
[CO2]eq=[CO2]o
4.5*10^-7=[H+][HCO3-]/[CO2]=x^2/[CO2]
or,[CO2]eq=x^2/(4.5*10^-7)=[HCO3-]eq/(4.5*10^-7)
[CO2]eq=[HCO3-]eq^2/(4.5*10^-7) ...................(2)
total alkalnity=[HCO3-] +2[CO32-] +[OH-] -[H+]=2.0*10^-3 mol/L
plugging in eqn (a),(b),(1) & (2) in the equation for total alkalinity,
total alkalnity=[HCO3-] +2(5.012e-4.)[HCO3-]. =2.0*10^-3 mol/L
or, [HCO3-]=1.997*10^-3 mol/L
[CO32-]eq=(5.012e-4.)[HCO3-].=(5.012e-4.)(2.0*10^-3 mol/L)=1.002*10^-6 mol/l
[H+]eq=[OH-]eq=10^-7 mol/L
[CO2]=[HCO3-]eq^2/(4.5*10^-7) =(1.997*10^-3 mol/L)^2/(4.5*10^-7)=8.862 mol/L
[CO2]eq=8.862*10^3 mol/L