In the qualitative analysis scheme, magnesium and nickel precipitate from solution upon the addition of sodium hydroxide. Once separated from the remaining cations by filtration or decanting, the solid mixture is acidified and warmed to dissolve magnesium and nickel cations back into solution. Adding ammonia creates a buffer solution. (Remember, HCl and NH3 makes for NH4+ cation.) The buffered solution should just be basic (say, pH = 8). Adding sodium hydrogen phosphate (Na2HPO4) precipitates magnesium as MgNH4PO4 (Ksp = 3 x 10-13).
Ksp = [Mg2+]*[NH3]*[HPO42-] = 3 x 10-13
Molar solubility = 6.7 x 10-5 M
a) Calculate the [NH4+] in a pH of 8.
b) Calculate the molar solubility of MgNH4PO4 in a solution with pH of 8.
To calculate the [NH4+] in a pH of 8:
Step 1: Recall that pH is a measure of the concentration of hydrogen ions (H+). In a basic solution, the concentration of hydroxide ions (OH-) is higher than the concentration of hydrogen ions (H+).
Step 2: Use the fact that water undergoes autoionization, producing equal concentrations of H+ and OH-. In pure water, the concentration of H+ and OH- is 1 x 10^-7 M.
Step 3: Since the solution is basic (pH = 8), the concentration of H+ is lower than 1 x 10^-7 M. Let's assume it is x.
Step 4: Use the fact that the product of the concentration of H+ and OH- is a constant (1 x 10^-14). Since [H+][OH-] = 1 x 10^-14, we can write ([H+])(1 x 10^-7) = 1 x 10^-14. Therefore, x(1 x 10^-7) = 1 x 10^-14.
Step 5: Solve for x: x = (1 x 10^-14)/(1 x 10^-7) = 1 x 10^-7 M.
Step 6: Since NH4+ forms as a result of the reaction of NH3 with H+, the concentration of NH4+ will be the same as the concentration of H+. Therefore, the [NH4+] in a pH of 8 is 1 x 10^-7 M.
To calculate the molar solubility of MgNH4PO4 in a solution with a pH of 8:
Step 1: Recall that molar solubility (S) represents the maximum concentration of a solute that can dissolve in a solvent at a given temperature.
Step 2: Use the solubility product constant (Ksp) expression for MgNH4PO4: Ksp = [Mg2+][NH4+][HPO42-] = 3 x 10^-13.
Step 3: Substitute the known values: 3 x 10^-13 = (molar solubility of MgNH4PO4 in M)[NH4+][NH3]^2[HPO42-].
Step 4: Since the [NH4+] is 1 x 10^-7 M (as calculated in part a), substitute it into the equation: 3 x 10^-13 = (molar solubility of MgNH4PO4 in M)(1 x 10^-7 M)([NH3]^2)([HPO42-]).
Step 5: The pH of the solution is 8, which indicates that the solution should be basic and contain NH3. However, the specific concentration of NH3 is not provided. Without that information, we cannot determine the molar solubility of MgNH4PO4.
In summary, the [NH4+] in a pH of 8 is 1 x 10^-7 M. However, the molar solubility of MgNH4PO4 in a solution with a pH of 8 cannot be determined without knowing the specific concentration of NH3.
a) To calculate the [NH4+] in a pH of 8, we need to make use of the Henderson-Hasselbalch equation for the ammonium/ammonia buffer system:
pH = pKa + log([A-]/[HA])
In this case, ammonia (NH3) is the base (A-) and ammonium (NH4+) is the acid (HA).
The pKa of the ammonium/ammonia buffer system is approximately 9.25.
Since we know the pH is 8, we can rearrange the Henderson-Hasselbalch equation:
8 = 9.25 + log([NH3]/[NH4+])
Subtracting 9.25 from both sides gives us:
-1.25 = log([NH3]/[NH4+])
To remove the logarithmic term, we can take the antilog of both sides:
10^(-1.25) = [NH3]/[NH4+]
Using a calculator, we find that 10^(-1.25) is approximately 0.056.
So:
0.056 = [NH3]/[NH4+]
To determine the [NH4+] concentration, we rearrange the equation:
[NH4+] = [NH3]/0.056
b) The molar solubility of MgNH4PO4 in a solution with a pH of 8 can be calculated using the given Ksp equation:
Ksp = [Mg2+][NH3][HPO42-] = 3 x 10^(-13)
Since the Ksp value is given, we can use it to find the molar solubility of MgNH4PO4.
Ksp = [Mg2+][NH3][HPO42-] = (6.7 x 10^(-5))^2[NH3]
Rearranging the equation, we have:
[NH3] = Ksp/[(6.7 x 10^(-5))^2]
Plugging in the given values, we get:
[NH3] = (3 x 10^(-13))/[(6.7 x 10^(-5))^2]
Using a calculator, we find that [NH3] is approximately 0.0084 M.
Therefore, the [NH4+] concentration in a pH of 8 is approximately 0.056 and the molar solubility of MgNH4PO4 in a solution with a pH of 8 is approximately 0.0084 M.