Molybdenum (Mo) is a heavy metal that can have severe health effects on humans and animals. Tailings from a mining operation in Colorado have polluted surface waters used by grazing animals with Mo. Based on the following information determine the concentration of the three Mo species in the stream. [Mo]T = 5.1x10^-4 M pH = 4.20
H2MoO4 (aq) = HMoO4^1- (aq) + H+ (aq) pKa1 = 4.0
HMoO4^1- (aq) = MoO4^2- (aq) + H+ (aq) pKa2=4.24
What does [Mo]T mean?
Have you determined fractions of a polyprotic acid such as H2A?
Here is the denominator which I will call D,
D = (H^+)^2 + k1(H^+) + k1k2. You know H^+ from the pH. You know k1 and k2. Solve for D and use it below.
fraction of H2A = (H^+)^2/D
fraction of HA^- = k1(H^+)/D
fraction of A^2- = k1k2/D
Then fraction of each specie times initial M = molarity of each specie.
Final check. Sum of species should equal Total at start. Post your work if you have questions about this.
To determine the concentration of the three Mo species in the stream, we need to consider the pKa values and the pH of the solution. Let's start by examining the chemical equilibria involved.
1. H2MoO4 (aq) ⇌ HMoO4^- (aq) + H+ (aq) (pKa1 = 4.0)
2. HMoO4^- (aq) ⇌ MoO4^2- (aq) + H+ (aq) (pKa2 = 4.24)
Given:
[Mo]T = 5.1x10^-4 M
pH = 4.20
We can determine the concentrations of the Mo species step-by-step:
Step 1: Determine the concentration of H2MoO4 (aq)
Since H2MoO4 (aq) is a weak acid, we can assume that it is almost completely dissociated at pH 4.20. Therefore, [H2MoO4] ≈ [H2MoO4]T.
[H2MoO4] ≈ 5.1x10^-4 M
Step 2: Determine the concentration of the first Mo species, HMoO4^- (aq)
Using the equilibrium equation from Step 1, we can set up an expression for the dissociation constant (Ka1) of H2MoO4:
Ka1 = [HMoO4^-][H+]/[H2MoO4]
Assuming negligible [HMoO4^-] compared to [H2MoO4], we can simplify the equation:
Ka1 ≈ [H+]
Since we know the pH (4.20), we can calculate the concentration of HMoO4^-:
[HMoO4^-] ≈ 10^(-pH) ≈ 10^(-4.20)
Step 3: Determine the concentration of the second Mo species, MoO4^2- (aq)
Using the equilibrium equation from Step 2, we can set up an expression for the equilibrium constant (Kw2) of HMoO4^-:
Kw2 = [MoO4^2-][H+]/[HMoO4^-]
Assuming negligible [MoO4^2-] compared to [HMoO4^-], we can simplify the equation:
Kw2 ≈ [H+]
Since we know the pH (4.20), we can calculate the concentration of MoO4^2-:
[MoO4^2-] ≈ 10^(-pH) ≈ 10^(-4.20)
In summary, the approximate concentrations of the three Mo species in the stream are:
[H2MoO4] ≈ 5.1x10^-4 M
[HMoO4^-] ≈ 10^(-4.20)
[MoO4^2-] ≈ 10^(-4.20)
To determine the concentration of the three Mo species in the stream, we need to consider the dissociation constants (pKa values) and the pH of the solution.
First, let's consider the dissociation of H2MoO4 (aq) into HMoO4^1- (aq) and H+ (aq). The pKa1 value for this reaction is given as 4.0.
Since the pH is 4.20, which is slightly higher than the pKa1 value, we can assume that most of the H2MoO4 has dissociated into HMoO4^1-. Therefore, the concentration of HMoO4^1- in the stream is nearly equal to the total concentration of Mo, [Mo]T = 5.1x10^-4 M.
Next, let's consider the further dissociation of HMoO4^1- (aq) into MoO4^2- (aq) and H+ (aq). The pKa2 value for this reaction is given as 4.24.
Since the pH is higher than the pKa2 value, we can assume that the equilibrium between HMoO4^1- and MoO4^2- lies more towards the MoO4^2- species.
To determine the concentrations of HMoO4^1- and MoO4^2- in the stream, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
For the reaction HMoO4^1- (aq) = MoO4^2- (aq) + H+ (aq), we can rewrite the equation as:
pH = pKa2 + log ([MoO4^2-]/[HMoO4^1-])
Plugging in the known values:
4.20 = 4.24 + log ([MoO4^2-]/[HMoO4^1-])
Rearranging the equation:
4.20 - 4.24 = log ([MoO4^2-]/[HMoO4^1-])
-0.04 = log ([MoO4^2-]/[HMoO4^1-])
To solve for the ratio [MoO4^2-]/[HMoO4^1-], we can take the antilog (-0.04):
[MoO4^2-]/[HMoO4^1-] = 10^(-0.04)
[MoO4^2-]/[HMoO4^1-] = 0.912
Therefore, the concentration of MoO4^2- in the stream is approximately 0.912 times the concentration of HMoO4^1-, which is equivalent to 0.912 times the total concentration of Mo, [Mo]T.
Concentration of MoO4^2- = 0.912 x [Mo]T
Concentration of HMoO4^1- = [Mo]T
Concentration of H2MoO4 = [Mo]T - Concentration of MoO4^2-