Consider a 1.50 L aqueous solution of 3.75 M NH3, where 17.5 g of NH4Cl are dissolved. To this solution, 5.00 g of MgCl2 are added.
a. What is [OH-] before MgCl2 is added?
b. Will a precipitate form?
c. What is [Mg2+] after equilibrium is established?
I don't need the answers, I just need an explanation of how to do these problems. Thanks.
That's good because I usually don't give answers anyway.
.........NH3 + H2O ==> NH4^+ + OH^-
I........3.75...........0.......0
C........-x.............x.......x
E......3.75-x...........x.......x
Kb = (NH4^+)(OH^-)/(NH3)
You want to solve for OH^-.
You know Kb.(NH4^+) = x from NH3 + ? from NH4Cl.
mols NH4Cl = grams/molar mass
M NH4Cl = mols/L solution. Total (NH4^+) = x + ?M NH4Cl
That's part a.
part b. Will a ppt form? That is will Mg(OH)2 ppt? If so Qsp must exceed Ksp.
Qsp = (Mg^2+)(OH^-)^2.
(Mg^2+) = M
mols MgCl2 = grams/molar mass
M MgCl2 = mols/L solution.
OH^- you know from above.
Solve for Qsp and compare with Ksp.
part c. I'm assuming Mg(OH)2 will ppt.
Ksp for Mg(OH)2 = (Mg^2+)(OH^-)^2
You know Ksp and (OH^-), plug those in and solve for (Mg^2+)
To solve these problems, we will use the principles of stoichiometry and equilibrium calculations. Here's how you can approach each question:
a. To determine [OH-] before MgCl2 is added, we need to consider the dissociation of NH3 in water. NH3 (ammonia) acts as a weak base and reacts with water to form NH4+ (ammonium) and OH- (hydroxide) ions. Since NH3 is a weak base, we assume that it does not significantly dissociate.
To calculate [OH-], we can use the expression for the equilibrium constant, known as the base dissociation constant (Kb), which relates the concentrations of NH4+ and OH- ions.
Kb = [NH4+][OH-] / [NH3]
The concentration of NH4+ can be calculated using the mass of NH4Cl and its molar mass, assuming complete dissociation:
moles of NH4+ = mass of NH4Cl / molar mass of NH4Cl
The concentration of NH3 can be calculated using its molarity and the volume of the solution:
moles of NH3 = molarity of NH3 x volume of solution
Finally, substituting these values into the Kb expression, we can solve for [OH-].
b. To determine if a precipitate will form, we need to consider the solubility of the products that may form during the reaction of the added MgCl2 with the NH3 in solution. We can look up the solubility rules to determine if a precipitate will form. If any insoluble product is formed, a precipitate will occur.
For example, if the reaction between MgCl2 and NH3 produces Mg(OH)2, we need to check its solubility. If Mg(OH)2 is insoluble, a precipitate will form.
c. To calculate the concentration of Mg2+ after equilibrium is established, we need to consider the reaction between MgCl2 and NH3. The balanced equation for this reaction shows that one mole of MgCl2 reacts with one mole of NH3 to form one mole of Mg(NH3)2Cl2.
Since we know the initial concentration of NH3 and the amount of MgCl2 added, we can determine the amount of Mg2+ formed based on the stoichiometry of the reaction. Then, by dividing the moles of Mg2+ by the final volume of the solution (after adding MgCl2), we can find the concentration of Mg2+.