What is the pH of a 0.40 mol L^-1 H2Se solution that has the stepwise dissociation constants
Ka1 = 1.3 × 10^-4 and Ka2 = 1.0 x 10^-11 ?
Thanks in advance
First, since the k2 is so small, ignore it. The pH will be determined by the dissociation of k2 only.
..........H2Se ==> H^+ + HSe^-
I.........0.4......0......0
C..........-x......x......x
E......0.4-x.......x......x
Substitue the E line into k1 expression and solve for x = (H^+), then convert to pH.
To determine the pH of the H2Se solution, we need to consider the stepwise dissociation of H2Se and the corresponding equilibrium expressions. H2Se dissociates as follows:
H2Se ⇌ H+ + HSe-
HSe- ⇌ H+ + Se2-
The dissociation constants given are for these two steps:
Ka1 = [H+][HSe-]/[H2Se]
Ka2 = [H+][Se2-]/[HSe-]
Given that the concentration of H2Se is 0.40 mol L^-1, we can set up an equilibrium table to calculate the concentrations of H+, HSe-, and Se2- at each step. Let's assume that the concentration of H+ from the dissociation of H2Se is x.
Step 1: H2Se ⇌ H+ + HSe-
[H2Se] = 0.40 M - x [H+] = x [HSe-] = x
Step 2: HSe- ⇌ H+ + Se2-
[HSe-] = 0.40 M - x [H+] = x [Se2-] = x
Now, let's use the given dissociation constants to set up the equilibrium expressions for each step:
Ka1 = (x)(x)/(0.40 - x)
Ka2 = (x)(x)/(0.40 - x)
Since the value of Ka1 (1.3 × 10^-4) is much greater than the value of Ka2 (1.0 × 10^-11), we can make the assumption that x << 0.40. This allows us to simplify the equilibrium expressions by neglecting the 'x' term in the denominator.
Now, we can solve the equation for Ka1:
1.3 × 10^-4 = (x)(x)/(0.40)
Cross-multiplying and rearranging the equation:
x^2 = (1.3 × 10^-4)(0.40)
x^2 = 5.2 × 10^-5
x = √(5.2 × 10^-5)
x ≈ 0.0072
So, the concentration of H+ is approximately 0.0072 M. To calculate the pH, we need to take the negative logarithm of the H+ concentration:
pH = -log[H+]
pH = -log(0.0072)
pH ≈ 2.14
Therefore, the pH of a 0.40 mol L^-1 H2Se solution with the given dissociation constants is approximately 2.14.