Calculate the concentrations of all species in a 1.67 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.
I know how to find the Na^+ and SO3^-2 but I don't know how to find molarity of rest of it. Can you show me how to find molarity please? Show me how to find molarity for HSO3^- and I will do the rest of them when I get it.
[Na^+] = 3.34M
[SO3^-2] = 1.67M
[HSO3^-] =
[H2SO3] =
[OH^-] =
[H^+] =
HSO3^- comes from the hydrolysis of SO3^2-.
..........SO3^2- + H2O ==> HSO3^- + OH^-
I.........1.67...............0.......0
C...........-x...............x.......x
E........1.67-x..............x.......x
Kb for SO3^2- = Kw/Ka2 for H2SO3 = (x)(x)/(SO3^2-)
Substitute and solve for x = (OH^-) = (HSO3^-)
I found all concentrations of all species except H2SO3. I keep getting wrong. The balanced equation is
H2SO3 ---> H^+ + HSO3
Is that correct?
I would look at this. Yes, your equation is right and I think if you substitute the values for H^+ and HSO3^- you will get H2SO3. An easier way may be this way. Try it and let me know
H2SO3 ==> 2H^+ + SO3^2- which isn't the way it ionizes BUT this can be used when one knows H^+ and SO3^2- as in this case.
k1k2 = 0.014*6.3E-8 = (H^+)^2(SO3^=)/(H2SO3) and solve for (H2SO3).
I finally got H2SO3 right.
H2SO3 ---> 2H^+ + SO3^-2 is work this way finding the H2S03. Thank you so much for your help. I appreciated your help.
To find the concentration of HSO3^- (bisulfite ion) in the solution, we need to consider the ionization of sulfurous acid (H2SO3).
The ionization reaction of H2SO3 can be written as follows:
H2SO3 ⇌ H+ + HSO3^- (Equation 1)
The ionization constant, Ka1, for this reaction is given as 1.4× 10–2.
The equilibrium expression for Equation 1 can be written as:
Ka1 = [H+][HSO3^-] / [H2SO3]
We can assume that the concentration of H2SO3 is very close to its original concentration since it is a weak acid. Therefore, we can approximate the concentration of H2SO3 as the initial concentration of Na2SO3, which is given as 1.67 M.
Now we can rearrange the equation to solve for [HSO3^-]:
[HSO3^-] = (Ka1 * [H2SO3]) / [H+]
Substituting the values we know:
[HSO3^-] = (1.4×10^-2 * 1.67 M) / [H+]
However, we don't have the concentration of [H+], which is the concentration of hydronium ions.
In order to determine the concentration of [H+], we need to consider the ionization of water.
H2O ⇌ H+ + OH^- (Equation 2)
The equilibrium constant for this reaction, Kw, is the product of the concentrations of [H+] and [OH^-], and has a constant value of 1.0 x 10^-14 at 25°C.
Therefore, we can write:
Kw = [H+][OH^-]
Since the solution is neutral (not acidic or basic), the concentration of [H+] is equal to the concentration of [OH^-]. Let's call this concentration x.
Therefore:
x * x = 1.0 x 10^-14
Solving for x:
x = sqrt(1.0 x 10^-14)
x ≈ 1.0 x 10^-7 M
Now we have the concentration of [H+] as 1.0 x 10^-7 M. We can substitute this value back into the equation for [HSO3^-]:
[HSO3^-] = (1.4×10^-2 * 1.67 M) / (1.0 x 10^-7 M)
[HSO3^-] ≈ 2.348 M
Therefore, the concentration of HSO3^- in the solution is approximately 2.348 M.
Now you can use this approach to find the concentrations of the remaining species in the solution.