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

Explanation:

The value of first dissociation constant;

$pK_{a1}=-\log[K_{a1}]$

$6.35=-\log[K_{a1}]$

$K_{a1}=4.467\times 10^{-7}$

The value of seconddissociation constant;

$pK_{a2}=-\log[K_{a2}]$

$10.33=-\log[K_{a2}]$

$K_{a2}=4.467\times 10^{-11}$

The pH of the water = 7.5

$pH=\log[H^+]$

$7.5=\log[H^+]$

$[H^+]=3.162\times 10^{-8} M$

$H_2CO_3\rightleftharpoons HCO_3^{-}+H^+$

 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:

$K_{a2}=\frac{y\times y}{(x-y)}$

$4.677\times 10^{-11}=\frac{y^2}{(x-y)}$..[1]

$x+y=[H^+]$

$x+y=3.162\times 10^{-8}$...[2]

Solving [1] and [2]:

x = $3.045\times 10^{-8} M$

y = $1.1702\times 10^{-9} M$

Expression of an first dissociation constant will be given as:

$K_{a1}=\frac{x\times x}{(C-x)}$

$4.467\times 10^{-7}=\frac{x^2}{(C-x)}$

$4.467\times 10^{-7}=\frac{(3.045\times 10^{-8} M)^2}{(C-(3.045\times 10^{-8} M))}$

Solving for C:

$C = 3.253\times 10^{-8} M$

At equilibrium , concentration of species:

Carbonic acid :

$[H_2CO_3]=(C-x)=3.253\times 10^{-8} M-3.045\times 10^{-8} M$

$[H_2CO_3]=2.08\times 10^{-9} M$

Carbonate ion :

$[CO_3^{2-}]=y=1.1702\times 10^{-9} M$

Bicarbonate :

$[HCO_3^{-}]=(x-y)=3.045\times 10^{-8} M-1.1702\times 10^{-9} M=2.928\times 10^{-8} M$Total carbonates:[TC]

$[TC]=[H_2CO_3]+[HCO_3^{-}]+[CO_3^{2-}]=C$

$= [TC} = 3.253\times 10^{-8} M$