Question

Question #dc4c4

Answer 1

`[[H_2CO_3]]/[[HCO_3^-]]=0.0823`

Explanation:

Let's start out by writing the balanced equation for the dissociation of carbonic acid:

`H_2CO_3rightleftharpoonsH^++HCO_3^-`

Remember the dissociation constant will be the ratio of the dissociated products (`H^+` and `HCO_3^-`) to the carbonic acid (`H_2CO_3`).

`K_d=([H^+][HCO_3^-])/[[H_2CO_3]]`

Looking at this equation, we can divide both sides by `[H^+]` to obtain a ratio between the bicarbonate and the carbonic acid.

`([HCO_3^-])/[[H_2CO_3]]=K_d/[[H^+]]`

Since we want the ratio of carbonic acid to bicarbonate, just invert the equation:

`([H_2CO_3])/[[HCO_3^-]]=[[H^+]]/K_d`

Now, all we need to do is find `[H^+]` using the definition of pH:

`pH=-log[H^+]`
`[H^+]=10^(-pH)`
`[H^+]=10^-7.45`
`[H^+]=3.55 *10^-8M`

Plugging back in, the ratio of carbonic acid to bicarbonate becomes:

`([H_2CO_3])/[[HCO_3^-]]=[[H^+]]/K_d`
`([H_2CO_3])/[[HCO_3^-]]=(3.55*10^-8)/(4.31*10^-7)`
`([H_2CO_3])/[[HCO_3^-]]=0.0823`

Answer 2

One can also use the Henderson-Hasselbalch Equation to calculate the Bicarb:Carbonic Acid Ratio then take the reciprocal.

Explanation:

`pH = pKa + log([HCO_3^-]/[H_2CO_3])`

`7.45 = -log(4.31E-7)` + `log([HCO_3^-]/[H_2CO_3])`

`7.45 = 6.37 + log([HCO_3^-]/[H_2CO_3])`

`1.085 = log([HCO_3^-]/[H_2CO_3])`

`([HCO_3^-]/[H_2CO_3])` = `10^(1.085) = 12.162`

`([H_2CO_3]/[HCO_3^-])` = `(1/12.162) = 0.0822`

`[H_2CO_3]:[HCO_3^-] = 0.0822 : 1.0000`