Calculate the concentrations of all species in a 1.17 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.
I found Na+ = 2.34
SO3^2- = 1.17
H2SO3 = 7.14 x10^-13
I can't find
HSO3^- = ?
OH^- = ?
H^+ = ?
Please show steps. THANK YOU!
Na is right.
The others are not. Na2SO3 is a salt and when it dissolves the SO3^2- hydrolyzes.
Na2SO3 ==> 2Na^+ + SO3^2-
1.17M.....2*1.17...1.17
.......SO3^2- + HOH --> HSO3^- + OH^-
I......1.17..............0........0
C.......-x...............x........x
E.....1.17-x.............x........x
Kb for SO3^2- = (Kw/Ka2 for H2SO3) = (x)(x)/(1.17-x)
Solve for x = OH^- = HSO3^-
I didn't solve this but the number is relatively small; I'll just call it y and use that below.
Then HSO3^- hydrolyzes to
...........HSO3^- + HOH ==> H2SO3 + OH^-
I...........y................0.......0
C...........-z...............z.......z
E...........y-z..............z.......z
Then
Kb for HSO3^- = (Kw/k1 for H2SO3) =
(z)(z)/(y-z).
You really don't need to solve this for the following reasons.
Look at Kb = Kw/k1 = 1E-14/1.4E-2 = about 1E-12 which is a very small number. So whatever you started with at y can subtract a small z and still essentially be y. Then look at z. That's so small coming from this hydrolysis that it's basically what you started with. See the first hydrolysis you calculated OH and that was the same as HSO3^-. So if you write the hydrolysis equation out like this
Kh = Kw/k1 = (H2SO3)(OH^-)/(HSO3^-) you see that OH^- = HSO3^- so those cancel each other and (H2SO3) = Kh = (Kw/k1) = about 1E-12 from above.
To find the concentrations of HS0₃⁻, OH⁻, and H⁺ in the solution, we need to consider the ionization reactions of sulfurous acid (H₂SO₃). There are two ionization steps:
1) H₂SO₃ ⇌ H⁺ + HSO₃⁻
2) HSO₃⁻ ⇌ H⁺ + SO₃²⁻
Let's start by calculating the concentration of HSO₃⁻:
For the first ionization step, we can use the equilibrium expression:
K₁ = [H⁺][HSO₃⁻] / [H₂SO₃]
Given that the initial concentration of H₂SO₃ is 1.17 M, and as the first ionization constant (Ka₁) is 1.4 × 10⁻², we can set up the equation:
1.4 × 10⁻² = [H⁺][HSO₃⁻] / 1.17
Rearranging the equation, we get:
[HSO₃⁻] = (1.4 × 10⁻²) × 1.17 / [H⁺]
To determine the concentration of H⁺, we need to consider both ionization reactions. Assuming x is the concentration of H⁺, we can write:
[H⁺] = x
For the second ionization step, we consider the equilibrium expression:
Ka₂ = [H⁺][SO₃²⁻] / [HSO₃⁻]
Knowing that Ka₂ is 6.3 × 10⁻⁸, [H⁺] = x, and [HSO₃⁻] is given by our previous equation, we can set up the equation:
6.3 × 10⁻⁸ = x [SO₃²⁻] / [(1.4 × 10⁻²) × 1.17 / x]
Simplifying the equation, we get:
6.3 × 10⁻⁸ = (x²)([SO₃²⁻]) / [(1.4 × 10⁻²) × 1.17]
Now, let's calculate the concentration of OH⁻. Since NaOH is a strong base and Na₂SO₃ is a salt, we can consider it to be fully dissociated in water. Thus, the concentration of OH⁻ will be twice that of Na₂SO₃.
OH⁻ concentration = 2 × [Na₂SO₃] = 2 × 1.17 M
Finally, considering that water is neutral, we can use the equation:
[H⁺][OH⁻] = 1.0 × 10⁻¹⁴
Substituting the values we obtained earlier, we have:
(H⁺)(2 × 1.17) = 1.0 × 10⁻¹⁴
2.34H⁺ = 1.0 × 10⁻¹⁴
Dividing both sides by 2.34, we get:
H⁺ ≈ 4.27 × 10⁻¹⁵ M
So, the concentrations of the species in the 1.17 M Na₂SO₃ solution are as follows:
HSO₃⁻ ≈ (1.4 × 10⁻²) × 1.17 / 4.27 × 10⁻¹⁵ M
OH⁻ ≈ 2 × 1.17 M
H⁺ ≈ 4.27 × 10⁻¹⁵ M