If the solubility of Ag2CO3 is 1.3x10^-4 mol/L, what is its Ksp?
(the answer is said to be 8.8x10^-12)
Ag2CO3(s) -----> 2 Ag+ (ag) + CO3 2-
Ksp= [Ag+]^2 [CO3^2-]
= [1.3x10^-4]^2[1.3x10^-4]
=2.197x10^-12
thats what I did, but I found that if you first multiplying the 1.3x10^-4 by 2 which gives you 2.6x10^-4 and then you insert it into the expression and multiply it by 1.3x10^-4 you get the answer. But it doesn't make sense to x2 it and then square it. Does anyone know why or if their answer is wrong and I'm right?
A typical student error. Yes, the answer is correct. For every 1 molecule of Ag2CO3 that dissolves you get TWICE the Ag^+ and ONE TIMES for the carbonate. So if Ag2CO3 is 1.3E-4 then Ag^+ is twice that and CO3^2- is 1.3E-4.
Well, you've stumbled upon a classic case of math and chemistry playing a little prank on us.
The correct expression for the solubility product constant (Ksp) is indeed Ksp = [Ag+]^2[CO3^2-]. However, to find the values that go into this expression, we need to consider the stoichiometry of the reaction.
In this case, the balanced equation shows that one Ag2CO3 yields two Ag+ ions. Therefore, if the solubility of Ag2CO3 is 1.3x10^-4 mol/L, we can say that the concentration of Ag+ ions is actually 2*(1.3x10^-4) = 2.6x10^-4 mol/L.
Substituting this value and the concentration of CO3^2- (1.3x10^-4 mol/L) into the expression, we get Ksp = (2.6x10^-4)^2 * (1.3x10^-4) = 8.8x10^-12, exactly as you mentioned.
So you were right all along! Multiplying the initial concentration of Ag2CO3 by 2 and then squaring it is the correct way to calculate the Ksp. It may seem a little counterintuitive, but when it comes to chemistry, sometimes you have to embrace the weirdness!
The correct expression for the solubility product constant (Ksp) of Ag2CO3 is:
Ksp = [Ag+]^2 [CO3^2-]
Given that the solubility of Ag2CO3 is 1.3x10^-4 mol/L, we can substitute this value into the expression:
Ksp = (1.3x10^-4)^2 (1.3x10^-4)
Simplifying this expression, we have:
Ksp = 1.69x10^-8 (1.3x10^-4)
Ksp = 2.197x10^-12
So, the correct answer is indeed 8.8x10^-12, not 2.197x10^-12.
Multiplying the solubility value by 2 (as you mentioned) is incorrect. The coefficient 2 in the chemical equation represents the number of moles of Ag+ ion produced for each mole of Ag2CO3 that dissolves. However, when calculating the Ksp, we are only concerned with the concentration of ions in the solution, not the stoichiometry of the reaction. Therefore, there is no need to multiply the solubility by 2 in this case.
The solubility product constant (Ksp) represents the equilibrium constant for the dissociation of a sparingly soluble compound in water. In the case of Ag2CO3, the equation is:
Ag2CO3(s) ⇌ 2 Ag+(aq) + CO3^2-(aq)
To calculate the Ksp, you need to know the solubility of Ag2CO3 in units of mol/L. The given solubility is 1.3x10^-4 mol/L.
Since Ag2CO3 dissociates into two Ag+ ions and one CO3^2- ion, the equilibrium expression for Ksp is:
Ksp = [Ag+]^2 [CO3^2-]
Substituting the given solubility, we have:
Ksp = (1.3x10^-4)^2 (1.3x10^-4)
= 2.197x10^-12
This is indeed the correct value for the Ksp of Ag2CO3.
It seems that you are confused about the step where you multiply the solubility by 2 before squaring it. You don't need to do this step because the stoichiometry of the dissociation reaction is already taken into account in the equilibrium expression. The coefficient "2" in front of Ag+ represents that there are two Ag+ ions formed for every molecule of Ag2CO3 that dissolves.
Therefore, in this case, multiplying the solubility by 2 (to account for the stoichiometry) and then squaring it would result in double counting the concentration of Ag+ ions. So, it is not correct to multiply the solubility by 2 before squaring it in the equilibrium expression.
Hence, the correct approach is to directly square the solubility of Ag2CO3 and multiply it by the solubility of CO3^2-, which is also 1.3x10^-4 mol/L, to obtain the Ksp value of 2.197x10^-12.