Question

Use the following equilibrium reaction and constant for the deprotonation of bicarbonate (HCO3-) to carbonate (CO32-) to determine: HCO3 = CO2 + H+ K = 10-10.33 (a) Whether HCO3 or CO32- would dominate at pH 9.1 and (b) What the concentration of [CO32-] would be at this pH if [HCO3 ] = 10-6 M

Answer 1

For a:
`HCO_3^-` will dominate at pH = 9.1

For b: The concentration of carbonate ions at pH = 9.1 will be
`5.9* 10^-8}M`

Step-by-step explanation:

To calculate the pH of the solution, we use the equation:

`pH=-\log[H^+]` ......(1)

• For a:

We are given:

pH = 9.1

Putting values in equation 1, we get:

`9.1=-\log[H^+]`

`[H^+]=10^(-9.1)`

For the given chemical equation:

`HCO_3^-\rightarrow CO_3^(2-)+H^+;K_a=10^(-10.33)`

The expression of
`K_a` for above reaction follows:

`K_a=([CO_3^(2-)]* [H^+])/([HCO_3^-])`

Putting value of hydrogen ion concentration in above equation, we get:

`10^(-10.33)=([CO_3^(2-)]* 10^(-9.1))/([HCO_3^-])\\\\([HCO_3^-])/([CO_3^(2-)])=(10^(-9.1))/(10^(-10.33))\\\\([HCO_3^-])/([CO_3^(2-)])=16.98`

`[HCO_3^-]=16.98* [CO_3^(2-)]`

Hence,
`HCO_3^-` will dominate at pH = 9.1

• For b:

The expression of
`K_a` for above reaction follows:

`K_a=([CO_3^(2-)]* [H^+])/([HCO_3^-])`

We are given:

`K_a=10^(-10.33)`

`[H^+]=10^(-9.1)`

`[HCO_3^-]=10^(-6)M`

Putting values in above equation, we get:

`10^(-10.33)=([CO_3^(2-)]* 10^(-9.1))/(10^(-6))`

`[CO_3^(2-)]=(10^(-6)* 10^(-10.33))/(10^(-9.1))=5.9* 10^-8}M`

Hence, the concentration of carbonate ions at pH = 9.1 will be
`5.9* 10^-8}M`