Respuesta :
Answer:
For a: [tex]HCO_3^-[/tex] will dominate at pH = 9.1
For b: The concentration of carbonate ions at pH = 9.1 will be [tex]5.9\times 10^-8}M[/tex]
Explanation:
To calculate the pH of the solution, we use the equation:
[tex]pH=-\log[H^+][/tex] ......(1)
- For a:
We are given:
pH = 9.1
Putting values in equation 1, we get:
[tex]9.1=-\log[H^+][/tex]
[tex][H^+]=10^{-9.1}[/tex]
For the given chemical equation:
[tex]HCO_3^-\rightarrow CO_3^{2-}+H^+;K_a=10^{-10.33}[/tex]
The expression of [tex]K_a[/tex] for above reaction follows:
[tex]K_a=\frac{[CO_3^{2-}]\times [H^+]}{[HCO_3^-]}[/tex]
Putting value of hydrogen ion concentration in above equation, we get:
[tex]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[/tex]
[tex][HCO_3^-]=16.98\times [CO_3^{2-}][/tex]
Hence, [tex]HCO_3^-[/tex] will dominate at pH = 9.1
- For b:
The expression of [tex]K_a[/tex] for above reaction follows:
[tex]K_a=\frac{[CO_3^{2-}]\times [H^+]}{[HCO_3^-]}[/tex]
We are given:
[tex]K_a=10^{-10.33}[/tex]
[tex][H^+]=10^{-9.1}[/tex]
[tex][HCO_3^-]=10^{-6}M[/tex]
Putting values in above equation, we get:
[tex]10^{-10.33}=\frac{[CO_3^{2-}]\times 10^{-9.1}}{10^{-6}}[/tex]
[tex][CO_3^{2-}]=\frac{10^{-6}\times 10^{-10.33}}{10^{-9.1}}=5.9\times 10^-8}M[/tex]
Hence, the concentration of carbonate ions at pH = 9.1 will be [tex]5.9\times 10^-8}M[/tex]
