What is the concentration of SO3^2- in a 0.043 M solution of Sulfurous acid (H2SO3) in water? The value of Ka1 for H2SO3 is 1.300 x 10-2 and the value of Ka2 is 6.300 x 10-8.
H2SO3 ==> H^+ + HSO3^-
HSO3^- ==> H^+ + SO3^2-
k1 is much larger than k2; therefore, H^+ is essentially equal to HSO3^-. So we write
k2 = (H^+)(SO3^2-)/(HSO3^-)
Since (H^+) = (HSO3^-), then
(SO3^2-) = k2.
To determine the concentration of SO3^2- in a 0.043 M solution of sulfurous acid (H2SO3) in water, we need to consider the dissociation of H2SO3 into its respective ions.
The chemical equation for the dissociation of H2SO3 can be written as follows:
H2SO3 ⇌ H+ + HSO3-
From the equation, we can see that for every one molecule of H2SO3 that dissociates, we obtain one H+ ion and one HSO3- ion.
Now let's calculate the concentration of H+ ions using the given Ka values. The Ka value for the dissociation reaction is given by:
Ka1 = [H+][HSO3-]/[H2SO3]
We are given the Ka1 value as 1.300 x 10-2. We can assume that the concentration of HSO3- is the same as that of H+ ions since they are formed in equal amounts. Therefore, we can write:
[H+]^2 / [H2SO3] = 1.300 x 10-2
Now, substitute the concentration of H2SO3, which is 0.043 M, into the equation:
[H+]^2 / 0.043 = 1.300 x 10-2
Rearranging the equation to solve for [H+], we have:
[H+]^2 = 1.300 x 10-2 * 0.043
[H+]^2 = 5.59 x 10-4
Taking the square root of both sides yields:
[H+] = √(5.59 x 10-4)
[H+] ≈ 0.0236 M
Since the concentration of H+ ions is equal to the concentration of HSO3- ions, the concentration of HSO3- in the solution is also 0.0236 M.
Finally, to find the concentration of SO3^2-, we need to consider the second dissociation step of HSO3-:
HSO3- ⇌ H+ + SO3^2-
The Ka2 value for this reaction is given as 6.300 x 10-8. Using a similar approach as before, we can write:
[H+]^2 / [HSO3-] = 6.300 x 10-8
Substituting the concentration of HSO3- (0.0236 M), we have:
[H+]^2 / 0.0236 = 6.300 x 10-8
Solving for [H+], we get:
[H+]^2 = 6.300 x 10-8 * 0.0236
[H+]^2 = 1.489 x 10-9
Taking the square root of both sides yields:
[H+] ≈ √(1.489 x 10-9)
[H+] ≈ 3.86 x 10-5 M
Thus, the concentration of SO3^2- in the solution is approximately 3.86 x 10-5 M.