Question

The pH of a water is measured to be 7.5. The system is open to atmosphere and the temperature is 25 oC. Assume that the system is in equilibrium with atmosphere, calculate the concentrations of carbonic acid, bicarbonate, carbonate, and CT (total carbonates). (given pKa1 and pKa2 of H2CO3 are 6.35 and 10.33, respectively).

Answer 1

Step-by-step explanation:

The value of first dissociation constant;

`pK_(a1)=-\log[K_(a1)]`

`6.35=-\log[K_(a1)]`

`K_(a1)=4.467* 10^(-7)`

The value of seconddissociation constant;

`pK_(a2)=-\log[K_(a2)]`

`10.33=-\log[K_(a2)]`

`K_(a2)=4.467* 10^(-11)`

The pH of the water = 7.5

`pH=\log[H^+]`

`7.5=\log[H^+]`

`[H^+]=3.162* 10^(-8) M`

`H_2CO_3\rightleftharpoons HCO_3^(-)+H^+`

C 0 0

At equilibrium

(C-x) x x

`HCO_3^(-)\rightleftharpoons CO_3^(2-)+H^+`

x 0 0

At equilibrium

(x -y) y y

Expression of an second dissociation constant will be given as:

`K_(a2)=(y* y)/((x-y))`

`4.677* 10^(-11)=(y^2)/((x-y))`..[1]

`x+y=[H^+]`

`x+y=3.162* 10^(-8)`...[2]

Solving [1] and [2]:

x =
`3.045* 10^(-8) M`

y =
`1.1702* 10^(-9) M`

Expression of an first dissociation constant will be given as:

`K_(a1)=(x* x)/((C-x))`

`4.467* 10^(-7)=(x^2)/((C-x))`

`4.467* 10^(-7)=((3.045* 10^(-8) M)^2)/((C-(3.045* 10^(-8) M)))`

Solving for C:

`C = 3.253* 10^(-8) M`

At equilibrium , concentration of species:

Carbonic acid :

`[H_2CO_3]=(C-x)=3.253* 10^(-8) M-3.045* 10^(-8) M`

`[H_2CO_3]=2.08* 10^(-9) M`

Carbonate ion :

`[CO_3^(2-)]=y=1.1702* 10^(-9) M`

Bicarbonate :

`[HCO_3^(-)]=(x-y)=3.045* 10^(-8) M-1.1702* 10^(-9) M=2.928* 10^(-8) M`Total carbonates:[TC]

`[TC]=[H_2CO_3]+[HCO_3^(-)]+[CO_3^(2-)]=C`

`= [TC} = 3.253* 10^(-8) M`