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

Answer:

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

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

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:

$K_a=\frac{[CO_3^{2-}]\times [H^+]}{[HCO_3^-]}$

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

$10^{-10.33}=\frac{[CO_3^{2-}]\times 10^{-9.1}}{[HCO_3^-]}\\\\\frac{[HCO_3^-]}{[CO_3^{2-}]}=\frac{10^{-9.1}}{10^{-10.33}}\\\\\frac{[HCO_3^-]}{[CO_3^{2-}]}=16.98$

$[HCO_3^-]=16.98\times [CO_3^{2-}]$

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

• For b:

The expression of $K_a$ for above reaction follows:

$K_a=\frac{[CO_3^{2-}]\times [H^+]}{[HCO_3^-]}$

We are given:

$K_a=10^{-10.33}$

$[H^+]=10^{-9.1}$

$[HCO_3^-]=10^{-6}M$

Putting values in above equation, we get:

$10^{-10.33}=\frac{[CO_3^{2-}]\times 10^{-9.1}}{10^{-6}}$

$[CO_3^{2-}]=\frac{10^{-6}\times 10^{-10.33}}{10^{-9.1}}=5.9\times 10^-8}M$

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