Calculate the concentrations of all species in a 0.690 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8.
.......SO3^2- + H2O ==> HSO3^- + OH^-
I......0.690M...........0.........0
C.......-x..............x.........x
E.....0.690-x...........x..........x
Kb1 for SO3^2- = (Kw/ka2 for H2SO3) = (HSO3^-)(OH^-)/(SO3^2-)
Solve for x = (OH^-) = (HSO3^-)
Then...HSO3^- + OH^- ==> H2SO3 + OH^-
Kb2 = (Kw/Ka1) = (H2SO3)(OH^-)/(HSO3^-)
BUT (HSO3^-) = (OH^-) from the first hydrolysis; therefore,, (H2SO3) = (Kw/Ka1).
All of that gives you H2SO3, HSO3^-, SO3^2-, OH^- and you can use (H^+)(OH^-)= Kw = 1E-14 to calculate H+.
To calculate the concentrations of all species in a 0.690 M Na2SO3 solution, we need to consider the dissociation of sodium sulfite (Na2SO3) in water.
The balanced equation for the dissociation of sodium sulfite can be written as follows:
Na2SO3(aq) ⇌ 2Na+(aq) + SO3^2-(aq)
Let's assume that x represents the concentration of SO3^2- in the solution when it dissociates. Therefore, the concentration of Na+ will also be 2x since two Na+ ions are produced for each SO3^2- ion.
Since sodium sulfite is a strong electrolyte, we can assume that it completely dissociates in water.
Using the given concentration of the sodium sulfite solution (0.690 M), we can set up an equation to find the concentration of SO3^2- (x).
0.690 = 2x
Solving for x, we find:
x = 0.345 M
Therefore, the concentration of SO3^2- in the solution is 0.345 M, and the concentration of Na+ is 2 * 0.345 M = 0.690 M.
Now, let's consider the ionization of the sulfurous acid (H2SO3):
H2SO3(aq) ⇌ H+(aq) + HSO3^-(aq)
The dissociation of sulfurous acid occurs in two steps because it has two hydrogen ions that can ionize.
Step 1: H2SO3(aq) ⇌ H+(aq) + HSO3^-(aq)
Using the given Ka1 for sulfurous acid (1.4×10^–2), we can assume that x represents the concentration of H+ and HSO3^- in the solution.
[H+] = [HSO3^-] = x
Therefore, [H2SO3(aq)] = 0.690 M - [H+] = 0.690 M - x
Since the initial concentration of H2SO3 is equal to its final concentration (before and after dissociation), we can set up the following equation:
Ka1 = [H+][HSO3^-] / [H2SO3(aq)]
1.4×10^–2 = x * x / (0.690 - x)
Solving for x, we find:
x ≈ 1.4×10^–2 M
Therefore, the concentration of H+ and HSO3^- in the solution is approximately 1.4×10^–2 M.
Now, let's move on to the second ionization step of sulfurous acid:
Step 2: HSO3^-(aq) ⇌ H+(aq) + SO3^2-(aq)
Using the given Ka2 for sulfurous acid (6.3×10^–8), we can again assume that x represents the concentration of H+ and SO3^2- in the solution.
[H+] = [SO3^2-] = x
Therefore, [HSO3^-] = 1.4×10^–2 M - [H+] ≈ 1.4×10^–2 M - x
Since the initial concentration of HSO3^- is equal to its final concentration (before and after dissociation), we can set up the following equation:
Ka2 = [H+][SO3^2-] / [HSO3^-]
6.3×10^–8 = x * x / (1.4×10^–2 - x)
Solving for x, we find:
x ≈ 2.5×10^–5 M
Therefore, the concentration of H+ and SO3^2- in the solution is approximately 2.5×10^–5 M.
In summary, the concentrations of all species in the 0.690 M Na2SO3 solution are as follows:
[Na+] = 0.690 M (unchanged from the initial concentration of Na2SO3)
[SO3^2-] = 0.345 M
[H+] = 1.4×10^–2 M
[HSO3^-] ≈ 1.4×10^–2 M
+ [SO3^2-] ≈ 2.5×10^–5 M
To calculate the concentrations of all species in a sodium sulfite (Na2SO3) solution, we need to consider the dissociation of sodium sulfite and the ionization of sulfurous acid (H2SO3) formed from the solute.
First, let's write down the balanced equation for the dissociation of sodium sulfite in water:
Na2SO3 (aq) → 2Na+ (aq) + SO3^2- (aq)
Since sodium sulfite fully dissociates, we can directly calculate the concentrations of Na+ and SO3^2- ions using the initial concentration of the sodium sulfite solution (0.690 M):
[Na+] = 2 * [Na2SO3] = 2 * 0.690 M = 1.380 M
[SO3^2-] = [Na2SO3] = 0.690 M
Next, let's consider the ionization of sulfurous acid (H2SO3) formed when sulfite ions react with water:
H2SO3 (aq) ⇌ H+ (aq) + HSO3^- (aq)
The concentration of H2SO3 can be calculated using the equilibrium expression:
Ka1 = [H+] * [HSO3^-] / [H2SO3]
We can approximate the concentration of H2SO3 as the concentration of SO3^2- (0.690 M) since sulfurous acid is formed from sulfite ions:
[H2SO3] ≈ [SO3^2-] = 0.690 M
Substituting the values into the equilibrium expression, we can solve for [H+]:
1.4× 10–2 = [H+] * [HSO3^-] / 0.690
Since the equilibrium constant Ka2 is much smaller than Ka1, we can assume that only a small fraction of the HSO3^- ion will further ionize:
[HSO3^-] ≈ [H2SO3]
Now we have:
1.4× 10–2 = [H+] * [H2SO3] / 0.690
Solving for [H+], we get:
[H+] ≈ 1.4× 10–2 * 0.690 / [H2SO3]
Note: The exact concentration of [H+] can be determined by solving a quadratic equation, but the approximation is usually accurate enough unless specifically requested.
Therefore, to calculate the concentration of [H+], we need the concentration of [H2SO3]. Since [H2SO3] ≈ [SO3^2-] = 0.690 M, we can use this value in the equation above.
Finally, to calculate the concentration of [HSO3^-], we can use the equation:
[HSO3^-] = [H2SO3] = 0.690 M
So, the concentrations of all species in the sodium sulfite solution are:
[Na+] = 1.380 M
[SO3^2-] = 0.690 M
[H2SO3] ≈ [SO3^2-] = 0.690 M
[HSO3^-] = [H2SO3] = 0.690 M
[H+] ≈ 1.4× 10–2 * 0.690 / [H2SO3] (where [H2SO3] ≈ 0.690 M)