Determine the pH of a 0.45M solution of Na2SO3 where Ka1 = 1.5 x 10^(-2) and Ka2 = 6.3 x 10^(-8).
To determine the pH of a solution of Na2SO3, we need to consider the hydrolysis reactions of the weak acid and base components of the salt. In this case, the hydrolysis reactions involve the ions from the sodium sulfite (Na2SO3) dissociating in water.
The dissociation of Na2SO3 in water is as follows:
Na2SO3 -> 2Na+ + SO3^2-
The sodium ion (Na+) is the conjugate acid of a strong base (NaOH) and is considered to be neutral in water, so it has no effect on the pH of the solution. However, the sulfite ion (SO3^2-) can react with water to form a weak acid.
The hydrolysis reaction of the sulfite ion is as follows:
SO3^2- + H2O -> HSO3- + OH-
This reaction produces both the bisulfite ion (HSO3-) and hydroxide ion (OH-). The hydroxide ion increases the concentration of hydroxide ions in the solution, making it more basic, which will affect the pH of the solution.
To find the pH of the solution, we need to calculate the concentrations of the bisulfite ion (HSO3-) and hydroxide ion (OH-) and use these values to determine the pH.
Step 1: Calculate the concentration of the bisulfite ion (HSO3-):
Since Na2SO3 dissociates completely in water, the concentration of the sulfite ion (SO3^2-) is equal to the initial concentration of Na2SO3. Therefore, the concentration of the bisulfite ion (HSO3-) is also equal to the initial concentration of Na2SO3.
[HSO3-] = 0.45 M
Step 2: Calculate the concentration of the hydroxide ion (OH-):
The hydroxide ion concentration ([OH-]) can be determined by using the equilibrium constant (Ka) for the hydrolysis reaction of the sulfite ion.
Ka2 = [HSO3-][OH-]/[SO3^2-]
Given that Ka2 = 6.3 x 10^(-8), and we already know the concentration of the bisulfite ion ([HSO3-]) is 0.45 M, we can rearrange the equation to solve for [OH-].
[OH-] = (Ka2 x [SO3^2-]) / [HSO3-]
Since the concentration of the sulfite ion ([SO3^2-]) is equal to the initial concentration of Na2SO3, we can substitute [SO3^2-] with 0.45 M.
[OH-] = (6.3 x 10^(-8) x 0.45) / 0.45
[OH-] = 6.3 x 10^(-8) M
Step 3: Calculate the pOH:
The pOH is the negative logarithm of the hydroxide ion concentration.
pOH = -log([OH-])
pOH = -log(6.3 x 10^(-8))
pOH ≈ 7.20
Step 4: Calculate the pH:
The pH of a solution is related to the pOH by the equation pH + pOH = 14.
pH + 7.20 = 14
pH ≈ 6.80
Therefore, the pH of the 0.45 M Na2SO3 solution is approximately 6.80.
To determine the pH of a solution of Na2SO3, we need to consider the dissociation of the sodium sulfite (Na2SO3) in water.
The compound Na2SO3 can be broken down into its ions in the following way:
Na2SO3 ⇌ 2Na+ + SO3^2-
First, we need to determine the concentration of the relevant ions in the solution:
[Na2SO3] = 0.45 M
[Na+] = 2 × 0.45 M = 0.9 M
[SO3^2-] = 0.45 M
Since sodium ions (Na+) do not contribute to the acidity of the solution, we will only consider the sulfite ions (SO3^2-) in calculating the pH.
Now, we need to recognize that Na2SO3 can undergo two-step dissociation, as indicated by the given Ka1 and Ka2 values.
Let's assume that x is the concentration of sulfite ions that dissociate in the first step. Thus, the concentration of sulfite ions remaining after the first dissociation is [SO3^2-] - x.
Using the first equilibrium reaction:
Ka1 = [H+][SO3^2-] / [Na2SO3]
Using the second equilibrium reaction:
Ka2 = [H+][HSO3^-] / [SO3^2-] - x
Since we are interested in the pH, we want to find the concentration of H+ ions.
For the first equilibrium, assume that x is small compared to the initial concentration of SO3^2-.
Ka1 = [H+][SO3^2-] / [Na2SO3]
1.5 × 10^(-2) = [H+][SO3^2-] / 0.45
Simplifying, [SO3^2-] = (1.5 × 10^(-2)) × (0.45) / [H+]
And for the second equilibrium, assume that the x value from the first step is small compared to [SO3^2-].
Ka2 = [H+][HSO3^-] / ([SO3^2-] - x)
6.3 × 10^(-8) = [H+][HSO3^-] / ([SO3^2-] - x)
Simplifying, [HSO3^-] = (6.3 × 10^(-8)) × ([SO3^2-] - x) / [H+]
Since we know that HSO3^- is protonated, HSO3^- behaves as a weak acid, and we can use the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log([HSO3^-] / [SO3^2-] - x)
Considering that the ionization constant of an acid (Ka) is equal to 10^(-pKa), we can substitute the values:
pH = -log(Ka2) + log([HSO3^-] / ([SO3^2-] - x))
Now, we can substitute the [HSO3^-] and [SO3^2-] expressions from the simplified equations earlier:
pH = -log(Ka2) + log((6.3 × 10^(-8)) × ([SO3^2-] - x) / [H+]) / ([SO3^2-] - x)
At this point, we can substitute the expression for [SO3^2-]:
pH = -log(Ka2) + log((6.3 × 10^(-8)) × ((1.5 × 10^(-2)) × (0.45) / [H+]) / [H+] - x)
Now, we can solve for x.
To determine x, we need to solve two equations simultaneously. The first equation comes from the statistical relationship:
[H+] = [HSO3^-] + [OH-]
The second equation comes from the water dissociation constant:
Kw = [H+][OH-]
Now, we can calculate the concentration of H+:
[H+] = (Kw / [SO3^2-]) - [SO3^2-]
Substituting the values:
[H+] = (10^(-14) / ([SO3^2-])) - ([SO3^2-])
By substituting [SO3^2-] = (1.5 × 10^(-2)) × (0.45) / [H+] into this equation, we can solve for [H+].
With the concentration of H+, we can substitute this value into the expression for x:
x = (6.3 × 10^(-8)) × ((1.5 × 10^(-2)) × (0.45) / [H+]) / [H+]
After calculating the value of x, we substitute it into the expression for pH, which we derived earlier:
pH = -log(Ka2) + log((6.3 × 10^(-8)) × ((1.5 × 10^(-2)) × (0.45) / [H+]) / [H+] - x)
By plugging in the values for the dissociation constants (Ka1 and Ka2), solving for x and substituting it back into the pH expression, you can determine the pH of the solution. However, due to the complexity of the calculations involved, it is recommended to use a specialized software or online calculator.
..................SO3^2- + HOH ==> HSO3^- + OH^-
I.................0.45..........................0..................0
C................-x.............................x...................x
E..............0.45-x........................x...................x
Kb1 for SO3^2- = (Kw/Ka2 for H2SO3) = (x)(x)/(0.45-x)
Solve for x = (OH^-) and convert to pH.
Post your work if you get stuck.