What is the PH of 0.40 M Na2SO3? (K1 for H2SO3 = 1.5 * 10^-2; K2 = 1.0 * 10^-7
.......SO3^2- HOH ==> HSO3^- + OH^-
I......0.4.............0........0
C......-x...............x.......x
E.....0.4-x.............x........x
Kb for SO3^2- = (Kw/K2 for H2SO3) = (HSO3^-)(OH^-)/(SO3^2-)
Substitute from the ICE chart, solve for x = OH^-, convert to pH.
To determine the pH of a solution, we need to consider the acidic or basic nature of the solute. In this case, we have a compound called sodium sulfite (Na2SO3), which can undergo hydrolysis and produce a weakly acidic solution due to the formation of sulfurous acid (H2SO3).
The sulfurous acid can dissociate in two stages:
Stage 1: H2SO3 ⇌ H+ + HSO3-
This equation has an equilibrium constant called K1 or Ka1, which is given as 1.5 * 10^-2.
Stage 2: HSO3- ⇌ H+ + SO3^2-
This equation has an equilibrium constant called K2 or Ka2, which is given as 1.0 * 10^-7.
To calculate the pH of the solution, we first need to determine the concentration of H+ ions formed from the hydrolysis of Na2SO3.
Since Na2SO3 contains two sodium ions (Na+) and one sulfite ion (SO3^2-), the concentration of each ion will be twice the concentration of Na2SO3.
[H+] = concentration of H+ ions
[HSO3-] = concentration of HSO3- ions
[SO3^2-] = concentration of SO3^2- ions
Therefore, [H+] = 2 * [Na2SO3]
Next, we'll use the equilibrium constant expression to calculate the concentration of H+ and HSO3- ions in the solution.
Ka1 = [H+][HSO3-]/[H2SO3]
[H2SO3] = [H+]/Ka1
Now, let's substitute the values into the equation:
[H2SO3] = [Na2SO3]/(2 * Ka1)
Similarly, we can calculate the concentration of HSO3- and SO3^2- ions:
[HSO3-] = [Na2SO3]/(2 * Ka1 * Ka2)
[SO3^2-] = [Na2SO3]/(2 * Ka1 * Ka2)
Finally, we can calculate the pH using the concentration of H+ ions:
pH = -log[H+]
Now, with the given values of [Na2SO3] and the equilibrium constants Ka1 and Ka2, you can plug in these values and calculate the pH of the 0.40 M Na2SO3 solution.
To find the pH of the solution, we need to determine the concentration of H+ ions in the solution.
Na2SO3 is a salt of a weak acid (H2SO3) and a strong base (NaOH), so we can assume that it undergoes hydrolysis in water. Hydrolysis reactions involve the splitting of water molecules, resulting in the formation of H+ or OH- ions.
The hydrolysis of Na2SO3 can be broken down into two steps:
1) Reacting with water to form NaOH, which is a strong base:
Na2SO3 + H2O → 2NaOH + H2SO3
2) Reacting with water to form H2SO3, which is a weak acid:
H2SO3 + H2O ⇌ H3O+ + HSO3-
Since K1 for H2SO3 is given as 1.5 × 10^-2, we can use this value to calculate the concentration of H+ ions in the solution.
Let x be the concentration of H+ ions in mol/L. Then, the concentration of HSO3- ions would also be x.
Since Na2SO3 completely dissociates into 2Na+ ions and one SO3^2- ion, the initial concentration of Na+ ions is equal to 2 times the concentration of Na2SO3 (0.40 M), which is 0.80 M.
Now, using the equilibrium constant expression for the hydrolysis of H2SO3, we have:
K1 = [H+][HSO3-] / [H2SO3]
Since the concentration of H2SO3 is not given, we can assume it to be negligible compared to the concentrations of H+ and HSO3- ions.
Therefore, we can simplify the equation to:
K1 = [H+][HSO3-]
Substituting x for both [H+] and [HSO3-]:
1.5 × 10^-2 = x²
Taking the square root of both sides:
x ≈ 0.1227 M
Now, we have the concentration of H+ ions in the solution. To convert this to pH, we can use the formula:
pH = -log[H+]
Substituting the value of [H+]:
pH = -log(0.1227)
pH ≈ 0.913
Therefore, the pH of a 0.40 M Na2SO3 solution is approximately 0.913.