A 0.24 M solution of the weak acid H2CO3 has a pH of 3.49. Determine the Ka for H2CO3 (carbonic acid).
Please show step by step.
Am i starting this correct?
H2CO3 + H2O -> H3CO3 + OH
or is it
H2CO3 + H2O -> HCO3 + H3O
Please show step by step.
Am i starting this correct?
H2CO3 + H2O -> H3CO3 + OH
or is it
H2CO3 + H2O -> HCO3 + H3O
-
The second equation is correct, since it is an acid, it donates a proton to form H3O or just H as it's usually written. Additionally, for an acid you don't need to put water in the equation. You do for a base however.
To start solving, set up an ICE table with the initial data.
For the concentrations of both HCO3- and H+, we can use the pH. Since they are both formed in equal amounts because of the 1:1 ratio, we can assume that they both have the same final concentrations.
Because pH is the -log([H+]), we can solve for [H+] by taking 10^-pH.
10^-3.49 = 3.24E-4, this is the final concentration of both HCO3 and H+.
H2CO3 -> HCO3- + H+
0.24 0 0
-x +x +x
0.24-x x x
Therefore, Ka = x^2/0.24-x
Since we know that x = 3.24E-4 (From the concentration of H+), we can substitute that into the equation to get:
Ka = (3.24E-4)^2/0.24 - 3.24E-4
Ka = 4.368E-7
To find the Kb, just divide it by Kw(1E-14)
To start solving, set up an ICE table with the initial data.
For the concentrations of both HCO3- and H+, we can use the pH. Since they are both formed in equal amounts because of the 1:1 ratio, we can assume that they both have the same final concentrations.
Because pH is the -log([H+]), we can solve for [H+] by taking 10^-pH.
10^-3.49 = 3.24E-4, this is the final concentration of both HCO3 and H+.
H2CO3 -> HCO3- + H+
0.24 0 0
-x +x +x
0.24-x x x
Therefore, Ka = x^2/0.24-x
Since we know that x = 3.24E-4 (From the concentration of H+), we can substitute that into the equation to get:
Ka = (3.24E-4)^2/0.24 - 3.24E-4
Ka = 4.368E-7
To find the Kb, just divide it by Kw(1E-14)