Calculate the pH of 0.2 M Na2SO3 data: ka1 (H2SO3) = 1.2 × 10 ^ (- 2) and Ka2 (H2SO3) = 6.2 × 10 ^ (- 8).
.............SO3^2- + HOH ==> HSO3^- + OH^-
I............0.2 M...............................0.............0
C.............-x...................................x..............x
E...........0.2-x.................................x...............x
Kb for SO3^2- = Kw/Ka for H2SO3 = x*x/0.2-x
Solve for x = OH^- and convert to pH.
Post your work if you get stuck.
To calculate the pH of a solution containing Na2SO3, we first need to understand the dissociation of the compound in water.
Na2SO3 dissociates in water to form sodium ions (Na+) and sulfite ions (SO3^2-). Since sulfite ions can react with water to form a weak acid, we need to consider the acidic dissociation equilibrium.
The sulfite ion (SO3^2-) can act as both a weak base and a weak acid. It reacts with water as follows:
SO3^2- + H2O ⇌ HSO3^- + OH-
The equilibrium constant for this reaction is given by Kb (HSO3^-) = [HSO3^-][OH-]/[SO3^2-], where [HSO3^-], [OH-], and [SO3^2-] are the concentrations of the respective ions at equilibrium.
Since we have values for the acid dissociation constants (Ka1 and Ka2) of the weak acid (H2SO3), which is formed by the reaction of sulfite ion with water, we can use these values to calculate the concentration of HSO3^- and OH- at equilibrium.
Let's assume that x is the concentration of HSO3^- and OH- ions at equilibrium. Then the concentration of SO3^2- at equilibrium is 0.2 - x (as Na2SO3 is a strong electrolyte and completely ionized).
Using the equilibrium expression, we can write:
Kb (HSO3^-) = x * x / (0.2 - x)
Now we can calculate the concentration of H+ ions by considering the second acidic dissociation equilibrium of H2SO3:
H2SO3 ⇌ H+ + HSO3^-
The equilibrium constant for this reaction is Ka2 (H2SO3) = [H+][HSO3^-]/[H2SO3].
Since the concentration of HSO3^- is x, the concentration of H+ ions will also be x. The concentration of H2SO3 is initially 0.2 M, but since it's a weak acid, we can consider the simplification that it dissociates negligibly.
So, [H+]/[H2SO3] ≈ x/0.2 = Ka2 (H2SO3)
Now we have an equation for the concentration of H+ ions in terms of x, which we obtained from the Kb expression. This lets us solve for x, the concentration of H+ ions.
Solving the quadratic equation of Kb using the given value of Kb of HSO3^-, we find that x = 2.08 × 10^(-4) M.
Therefore, the concentration of H+ ions (and hence the pH) in the Na2SO3 solution is 2.08 × 10^(-4) M.