Groundwater in Pherric, New Mexico, initially contains 1.800 mg/L of iron as Fe3+. What must the pH be raised to in order to precipitate all but 0.3 mg/L of the iron? The temperature of the solution is 25 degree Celsius.
Fe(OH)3 ==> Fe^+3 + 3OH^-
Ksp = (Fe^+3)(OH^-)^3
Set (Fe^+3) = 0.3 mg/L converted to moles/L and solve for OH^-. Then convert OH to pOH and to pH.
To determine the pH necessary to precipitate all but 0.3 mg/L of iron (Fe3+), we can use the solubility product constant (Ksp) of iron hydroxide (Fe(OH)3).
1. Write the balanced equation for the precipitation reaction:
Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH-(aq)
2. Write the expression for the solubility product constant (Ksp):
Ksp = [Fe3+][OH-]^3
3. Calculate the concentration of OH- ions needed to precipitate all but 0.3 mg/L of iron:
Fe3+ concentration = 1.800 mg/L (given)
OH- concentration = [Fe3+] / 3 (stoichiometric ratio)
OH- concentration = 1.800 mg/L / 3 = 0.600 mg/L
4. Convert the OH- concentration to molarity (M):
1 mg/L = 1 ppm (parts per million)
0.600 mg/L = 0.600 ppm = 0.600 mg/L / 1000 = 0.000600 g/L
molar mass of OH- = 17.01 g/mol
OH- concentration (M) = 0.000600 g/L / 17.01 g/mol = 3.527 × 10^(-5) M
5. Use the pH equation to find the pH required to achieve the desired OH- concentration:
pOH = -log[OH-]
pOH = -log(3.527 × 10^(-5))
pOH = 4.454
6. Calculate the pH:
pH + pOH = 14 (at 25°C)
pH = 14 - 4.454
pH = 9.546
Therefore, the pH must be raised to approximately 9.546 in order to precipitate all but 0.3 mg/L of the iron.
To determine the pH at which the iron precipitates in this scenario, we need to consider the solubility equilibrium of iron hydroxide (Fe(OH)3), which is the solid phase at which the iron will precipitate.
The solubility product constant (Ksp) expression for Fe(OH)3 at 25 °C is:
Ksp = [Fe3+][OH-]^3
Since the concentration of OH- is determined by the pH, we can rewrite the Ksp expression as:
Ksp = [Fe3+][(10^-pOH)^3]
Given the initial concentration of Fe3+ as 1.800 mg/L, we need to find the pH at which the concentration of Fe3+ drops to 0.3 mg/L. This means we want to determine the pH at which [Fe3+] = 0.3 mg/L.
Now let's calculate the concentration of Fe3+ in terms of molarity:
Molar mass of Fe = 55.845 g/mol
1 mg = 1/1000 g
Concentration of Fe3+ in mg/L = 1.800 mg/L = 1.800 mg/L * (1 g/1000 mg) = 0.001800 g/L
Concentration of Fe3+ in mol/L = 0.001800 g/L * (1 mol/55.845 g) = 3.22710^-5 mol/L
To find the pH at which [Fe3+] = 0.3 mg/L, we use the equation:
Ksp = [Fe3+][(10^-pOH)^3]
Rearranging the equation to solve for pOH:
pOH = -log((Ksp / [Fe3+])^(1/3))
Substituting the values:
Ksp = solubility product constant = 2.79 x 10^-39 (at 25 °C)
[Fe3+] = 0.3 mg/L = 0.3 mg/L * (1 g/1000 mg) * (1 mol/55.845 g) = 5.39210^-6 mol/L
pOH = -log((2.79 x 10^-39 / 5.39210^-6)^(1/3))
Calculating the pOH:
pOH = 38.98
To find the pH, we can use the relationship:
pH + pOH = 14
pH + 38.98 = 14
pH = 14 - 38.98
pH = -24.98
The pH needs to be raised to approximately -24.98 to precipitate all but 0.3 mg/L of the iron in the groundwater in Pherric, New Mexico. However, please note that a pH of -24.98 is not physically meaningful in the context of aqueous solutions. Typically, pH values range from 0 to 14, with 7 being neutral. So it's not possible to achieve the desired level of iron precipitation with a normal pH scale.