Calculate the concentrations of H2SO3, H3O+, HSO3-, and SO32- in a 0.025 M H2SO3(aq) solution
Look up k1 and k2 for H2SO3. k2 is so much smaller than k1 that the majority of the H^+ will be furnished by k1 and "almost" none by k2.
.........H2SO3 ==> H^+ + HSO3^-
initial.0.025M......0......0
change...-x..........x......x
equil....?..........x........x
k1 = (H^+)(HSO3^-)/(H2SO3)
Substitute and solve for x = (H^+)=(HSO3^-)
...........HSO3^- ==> H^+ + SO3^2-
Note from above that (H^+) = (HSO3^-)
k2 = (H^+)(SO3^2-)/(HSO3^-)
Since (H^+) = (HSO3^-) (almost anyway), then (SO3^2-) = k2.
To calculate the concentrations of H2SO3, H3O+, HSO3-, and SO32- in a 0.025 M H2SO3(aq) solution, we need to consider the dissociation of H2SO3 in water.
The balanced equation for the dissociation of H2SO3 in water is:
H2SO3 + H2O ⇌ H3O+ + HSO3-
From this equation, we can determine the equilibrium expression:
K = [H3O+][HSO3-] / [H2SO3]
Given that [H2SO3] = 0.025 M, we need to calculate the concentrations of H3O+, HSO3-, and SO32-.
Since H2SO3 is a weak acid, we can assume that its dissociation is minimal, so [H2SO3] is approximately equal to [H3O+]. Therefore, [H3O+] ≈ 0.025 M.
Since the dissociation of H2SO3 produces equal amounts of H3O+ and HSO3-, [HSO3-] ≈ 0.025 M.
Since both HSO3- and SO32- are formed from the dissociation of H2SO3, their concentrations will be equal, assuming the reaction goes to completion. Therefore, [SO32-] ≈ 0.025 M.
So, the concentrations are approximately:
[H2SO3] ≈ 0.025 M
[H3O+] ≈ 0.025 M
[HSO3-] ≈ 0.025 M
[SO32-] ≈ 0.025 M
To calculate the concentrations of different species in a solution, you need to consider the acid-base equilibrium reactions that occur in the solution. In this case, the relevant acid-base equilibria are:
1. H2SO3 ⇌ H+ + HSO3-
2. HSO3- ⇌ H+ + SO32-
Given:
Initial concentration of H2SO3 = 0.025 M
Let's denote the concentrations of H2SO3, H3O+, HSO3-, and SO32- by [H2SO3], [H3O+], [HSO3-], and [SO32-], respectively.
The initial concentration of H2SO3 is already given as 0.025 M, so [H2SO3] = 0.025 M.
Now, let's assume that at equilibrium, the concentration of H+ is x in both equations.
For the first equation:
[H2SO3] = [H+] + [HSO3-]
0.025 M = x + [HSO3-] (Equation 1)
For the second equation:
[HSO3-] = [H+] + [SO32-]
[HSO3-] = x + [SO32-] (Equation 2)
Since both equations (Equation 1 and Equation 2) involve a common species (H+), we can equate their right-hand sides:
x + [HSO3-] = [HSO3-] + [SO32-]
This simplifies to:
x = [SO32-]
Now, substituting the value of x (= [SO32-]) into Equation 1, we get:
0.025 M = [H+] + [HSO3-]
Since [H+] = x, we can rewrite the equation as:
0.025 M = x + [HSO3-]
Since we know that x = [SO32-], we can rewrite the equation as:
0.025 M = [SO32-] + [HSO3-] (Equation 3)
Now, in order to determine the concentrations of [H3O+], [HSO3-], and [SO32-], we need one more equation. This equation can be obtained by considering the water auto-ionization reaction:
H2O ⇌ H+ + OH-
In neutral solutions, the concentration of H+ ions is equal to the concentration of OH- ions, so [H+] = [OH-]. Assuming the concentration of OH- is y, we can write the equation as:
[H+] = y
Now, applying the expression for the equilibrium constant (Kw) for the water auto-ionization reaction, which is Kw = [H+][OH-] = 1.0 × 10^-14 at 25°C, we substitute [H+] with y:
y * y = 1.0 × 10^-14
Taking the square root of both sides, we get:
y = √(1.0 × 10^-14)
y ≈ 1.0 × 10^-7 M
Since [H+] = [OH-] ≈ 1.0 × 10^-7 M, we know the concentration of H3O+ ([H3O+]) is approximately 1.0 × 10^-7 M.
Now, substituting the calculated value of [H3O+] (= 1.0 × 10^-7 M) into Equation 3, we can solve for [HSO3-] and [SO32-]:
0.025 M = [SO32-] + [HSO3-]
Since [H3O+] = [SO32-] = 1.0 × 10^-7 M:
0.025 M = 1.0 × 10^-7 M + [HSO3-]
Simplifying the equation:
[HSO3-] ≈ (0.025 - 1.0 × 10^-7) M
Therefore, the approximate concentrations are:
[H2SO3] = 0.025 M
[H3O+] ≈ 1.0 × 10^-7 M
[HSO3-] ≈ (0.025 - 1.0 × 10^-7) M
[SO32-] ≈ 1.0 × 10^-7 M