The equilibrium constant for the reaction Cu3(PO4)2(s) 3 Cu2+(aq) + 2 PO43–(aq) is K = 1.40 x 10–37. Suppose that a beaker is filled to 1 L with a solution containing
[Cu2+] = [PO43–] = 10.0 nM, and 300.0 g Cu3(PO4)2 . Which of the following observations is true?
a.
Q < K, and products must be converted to reactants for the reaction to reach
equilibrium.
b.
Q < K, and reactants must be converted to products for the reaction to reach
equilibrium.
c.
Q > K, and products must be converted to reactants for the reaction to reach
equilibrium.
d.
Q > K, and reactants must be converted to products for the reaction to reach
equilibrium.
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Ok, So my first thought is to write the expression for Q:
Q = [Cu+2]^3[PO4-3]^2
Substitute 10.0nM --> 10 x 10^-9
Q= [10 x 10^-9]^5
= 1 x 10^-40
I don't know what to do with the 300.0g Cu3(PO4)2. Help please.
So far you are good. So Q = 1E-40 < K = 1.4E-37 which lets you know c and d can't be right. The answer must be a or b. Now write the Keq expression.
Keq (which is really Ksp) =
Cu3(PO4)2 = (Cu^2+)^3(PO4^3-)^2 = 1.4E-37
Q must be larger. How can it get larger? No way except for Cu^2+ and PO4^3- to increase so some of the solid MUST dissolve in order to increase the ion concentrations. What did I do with the 300 g Cu3(PO4)2. I said, aha, so it has SOME solid (in order for Ksp to be a Ksp there must be at least a speck of solid in the beaker and 300 g qualifies as a speck). Another way of saying this is that the ions concentrations are not high enough so some of the solid must dissolve in order to reach the equilibrium state. So the answer is ........?
thank you :)
To determine what to do with the 300.0 g of Cu3(PO4)2, we need to consider the stoichiometry of the reaction.
From the balanced equation:
Cu3(PO4)2(s) → 3 Cu2+(aq) + 2 PO43–(aq)
We can see that for every 1 mole of Cu3(PO4)2, we get 3 moles of Cu2+ ions and 2 moles of PO43– ions.
First, let's calculate the moles of Cu3(PO4)2:
Molar mass of Cu3(PO4)2 = (3 x molar mass of Cu) + (2 x molar mass of PO4)
= (3 x 63.55 g/mol) + (2 x (31.0 g/mol + 4 x 16.0 g/mol))
= 285.4 g/mol
Moles of Cu3(PO4)2 = mass / molar mass
= 300.0 g / 285.4 g/mol
= 1.05 mol of Cu3(PO4)2
Now, we can calculate the concentrations of Cu2+ and PO43– ions in the solution:
Concentration of Cu2+ and PO43– = moles / volume
= 1.05 mol / 1 L
= 1.05 M
Now, let's calculate the value of Q:
Q = [Cu2+]^3 * [PO43–]^2
= (1.05)^3 * (1.05)^2
= 1.05^5
≈ 1.276
Comparing Q to the given K value (1.40 x 10^–37), we can see that Q > K.
Therefore, the correct answer is c. Q > K, and products must be converted to reactants for the reaction to reach equilibrium.
To determine what to do with the 300.0 g of Cu3(PO4)2, we need to calculate the concentration of Cu3(PO4)2 in the solution.
We can start by finding the molar mass of Cu3(PO4)2.
The molar mass of Cu is 63.55 g/mol, and the molar mass of PO4 is 94.97 g/mol.
So, the molar mass of Cu3(PO4)2 would be:
(3 * 63.55 g/mol) + (2 * 94.97 g/mol) = 285.15 g/mol
Now, we can calculate the number of moles of Cu3(PO4)2 in 300.0 g:
moles of Cu3(PO4)2 = mass / molar mass
moles of Cu3(PO4)2 = 300.0 g / 285.15 g/mol
This will give us the number of moles of Cu3(PO4)2 in the solution.
Next, we need to calculate the volume of the solution in liters.
Since we have a 1 L beaker filled with solution, the volume is already given as 1 L.
With the concentration and volume, we can calculate the concentration of Cu3(PO4)2 in the solution.
Concentration = moles of Cu3(PO4)2 / volume
Substituting the values we have:
Concentration = (300.0 g / 285.15 g/mol) / 1 L
Simplifying gives us:
Concentration = 1.051 M
Now, we can use this concentration to calculate the new value of Q:
Q = [Cu2+]^3[PO43-]^2
Q = (1.051 M)^3 * (10 x 10^-9)^2
Calculating this value will give us the new value of Q.
Compare this value of Q to the given value of K (1.40 x 10^-37), and based on the comparison, you can determine the correct observation from the given options (a, b, c, d).