Calculate the concentrations of all species in a 1.41 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.
Na^+ = 2.82M
SO3^2- = 1.40M
HSO3^- = 0.00047M
H2SO3 =
OH^- =
H^+ =
I have been able to calculate the first three, but do not know where to go from there to find the last three. Any help would be very much appreciated.
I agree with Na^+ and HSO3^-. I would have used 1.41 M for SO3^2- since 1.41-0.00047 = essentially 1.41 M.
You must have written this equation to obtain the 0.00047.
SO3^2- + HOH ==> HSO3^- + OH^-
So (HSO3^-) = 0.00047 which makes OH^- = 0.00047 M.
For H^+, you know OH and
(H^+)(OH^-) = Kw which gives you H^+.
Finally, for H2SO3, you have the second hydrolysis of HSO3^- as
HSO3^- + HOH ==> H2SO3 + OH^-
Kb2 = (Kw/k2) = 1E-14/1.4E-2 and
1.4E-2 = (H2SO3)(OH^-)/(HSO3^-).
However, in the first hydrolysis (kb1) you said (HSO3^-) = (OH^-) so Kb2 = (H2SO3) = about 7.1E-13 M.
YOLO
How did you get 0.00047?
To calculate the concentrations of H2SO3 (sulfurous acid), OH- (hydroxide ions), and H+ (hydrogen ions), we need to consider the ionization of sulfurous acid.
The ionization of sulfurous acid (H2SO3) can be represented by the following equations:
H2SO3 ⇌ H+ + HSO3- (1)
HSO3- ⇌ H+ + SO3^2- (2)
Let's denote the initial concentration of H2SO3 as x. The equilibrium concentration of HSO3- can be assumed to be negligible compared to the initial concentration of H2SO3, so the concentration of H+ from equation (1) will be equal to the initial concentration of H2SO3.
Thus, we have [H+] = [H2SO3] = x.
Now, let's consider equation (2) where HSO3- is the acid and SO3^2- is the base. Since we know the initial concentration of H2SO3 (x) and the initial concentration of HSO3-, we can use the ionization constant Ka2 to calculate the concentration of SO3^2- at equilibrium.
Ka2 = [H+][SO3^2-] / [HSO3-]
6.3x10^-8 = x * [SO3^2-] / (0.00047M)
Rearrange the equation to solve for [SO3^2-]:
[SO3^2-] = (6.3x10^-8)* (0.00047M) / x
Now we have the concentration of [SO3^2-]. Since Na2SO3 is a strong electrolyte and completely dissociates into its ions, the concentration of Na+ will be twice the concentration of Na2SO3: [Na+] = 2 * 1.41 M = 2.82 M.
Now, using the charge balance, we can determine the concentration of OH- and H+ in the solution.
OH- + H+ = 14 pH = 14 pOH
14 - pH = pOH
For a solution, pH + pOH = 14.
Therefore, pH = 14 - pOH
Since we know that [H+][OH-] = 1x10^-14 (at 25°C), we can use our calculated value for [H+] to determine [OH-]:
[OH-] = 1x10^-14 / [H+]
Finally, we know that [H2SO3] = [H+], so we can directly substitute the calculated value of [H2SO3] or [H+] into the formulas for [OH-] and [H+].
Therefore, using the calculations you already made:
H2SO3 = [H+] = x (The initial concentration of H2SO3)
OH- = 1x10^-14 / [H+] = 1x10^-14 / x
H+ = [H2SO3] = x
So, to find the concentrations of H2SO3, OH-, and H+ in the solution, you need to substitute the value of [H2SO3] (or [H+]) that you calculated earlier into the formulas for OH- and H+.