What is the molar solubility of silver carbonate Ag2CO3 at 25 C if it is dissolved in a 0.15M solution of silver nitrate, AgNO3?
Solubility constant:
silver carbonate = 8.1 *10^(-12)
Answer: 3.6 *10^(-10)
So... not sure exactly what to do here.
AgCO3 + AgNO3 --><--
AgCO3 + Ag(+) + NO3(-) (?)
not sure what molarity has to do with this.
Any help would be appreciated.
Also
what is Ksp expression of mercury(I) chloride Hg2Cl2 in terms of molar solubility, s?
Answer: 4s^3
Why is this not s^2? I thought that mercury(I) was a diatomic and Cl2 had a -2 charge (and is also diatomic). That would mean
1 mercury(I)ion:1 chlorine ion or
s*s=s^2
The 0.15M has a lot to do with it. You have Ag^+ from two sources.
Ag2CO3 ==> 2Ag^+ + CO3^2- AND
AgNO3 ==> Ag^+ + NO3^-
Ksp = (Ag^+)^2(CO3^2-) = 8.1E-12
If you want to go through it rigorously, then Ag^+ from Ag2CO3 is 2x and CO3^2- is x and Ag^+ from AgNO3 (it is soluble and 100% ionized) is 0.15. Substituting we get
(2x+0.15)^2(x) = 8.1E-12.
The easy way to solve this is to recognize that the solubility of Ag2CO3 is small so 2x+0.15 is essentially equal to 0.15. The equation then becomes
(0.15)^2(x)=8.1E-12 and solve for x. That gives you the CO3^2- which ALSO is Ag2CO3.
For your second question, note the formula is Hg2Cl2 and NOT HgCl2.(mercury(I) chloride is a dimer.) so
Hg2Cl2(s) ==> Hg22+ + 2Cl^-
Ksp = (Hg22+)(Cl^-)^2
So Hg22+ = x and Cl^- is 2x. Substituting gives 4s^3
To find the molar solubility of silver carbonate (Ag2CO3) in a 0.15M solution of silver nitrate (AgNO3) at 25 degrees Celsius, you can use the solubility product constant (Ksp) and an ICE (Initial, Change, Equilibrium) table.
First, write the balanced chemical equation for the dissolution of silver carbonate in silver nitrate:
Ag2CO3 (s) ⇌ 2Ag+ (aq) + CO3^2- (aq)
Next, set up the ICE table:
Initial:
[Ag2CO3] = 0 (since it's a solid)
[Ag+] = 0.15 M (from the silver nitrate solution)
[CO3^2-] = 0 (since it hasn't dissolved yet)
Change:
[Ag2CO3] decreases by x (since it is dissolving)
[Ag+] increases by 2x (according to the stoichiometry of the balanced equation)
[CO3^2-] increases by x (according to the stoichiometry of the balanced equation)
Equilibrium:
[Ag2CO3] = x
[Ag+] = 0.15 + 2x
[CO3^2-] = x
Now, write the expression for the solubility product constant (Ksp):
Ksp = [Ag+]^2[CO3^2-] = (0.15 + 2x)^2(x)
Since the Ksp value for silver carbonate is given as 8.1 * 10^(-12), you can substitute this into the equation:
8.1 * 10^(-12) = (0.15 + 2x)^2(x)
To solve for x (molar solubility), you can use numerical methods such as trial and error or solve it algebraically. Once you find the value of x, that will give you the molar solubility of silver carbonate in the given silver nitrate solution.
As for the Ksp expression of mercury(I) chloride (Hg2Cl2) in terms of molar solubility (s), it is indeed (s^2) because Hg2Cl2 dissociates into Hg2^2+ and 2Cl^- ions. Each Hg2Cl2 molecule produces two Hg2^2+ ions and two Cl^- ions, so the Ksp expression becomes (4s^3). The square (s^2) would only be applicable if the dissociation resulted in one Hg2^2+ ion and one Cl^- ion.
To find the molar solubility of silver carbonate (Ag2CO3) in a 0.15M solution of silver nitrate (AgNO3), we can use the solubility product constant (Ksp) and the stoichiometry of the balanced equation.
First, let's write the balanced equation for the dissociation of silver carbonate in water:
Ag2CO3(s) ⇌ 2Ag+(aq) + CO3^2-(aq)
Given that the solubility constant (Ksp) for silver carbonate is 8.1 * 10^(-12), we can set up the expression for the equilibrium constant as follows:
Ksp = [Ag+]^2 [CO3^2-]
Since the solution contains a 0.15M solution of silver nitrate, we assume that the concentration of Ag+ is also 0.15M. However, at equilibrium, the concentration of Ag+ resulting from the dissociation of Ag2CO3 will be 2 times the molar solubility (s).
Therefore, we can rewrite the Ksp expression using the molar solubility (s):
8.1 * 10^(-12) = (2s)^2 [CO3^2-]
To solve for the molar solubility (s), we rearrange the equation and plug in the known value for Ksp:
s^2 = (8.1 * 10^(-12)) / (4)
s^2 = 2.025 * 10^(-12)
s = √(2.025 * 10^(-12))
s ≈ 1.425 × 10^(-6) M
Therefore, the molar solubility of silver carbonate (Ag2CO3) in a 0.15M solution of silver nitrate (AgNO3) at 25°C is approximately 1.425 × 10^(-6) M.
Regarding your second question about the Ksp expression of mercury(I) chloride (Hg2Cl2) in terms of molar solubility (s), the stoichiometry of the balanced equation plays a crucial role.
The balanced equation for the dissociation of mercury(I) chloride in water is:
Hg2Cl2(s) ⇌ Hg2^2+(aq) + 2Cl^-(aq)
In this case, the molar solubility (s) represents the concentration of Hg2^2+ ions, as there are two Hg2^2+ ions formed for each Hg2Cl2 molecule that dissociates.
The Ksp expression is given as:
Ksp = [Hg2^2+][Cl^-]^2
Since the concentration of Cl^- ions is 2 times the molar solubility (s) due to the stoichiometry of the balanced equation, we can rewrite the Ksp expression as:
Ksp = (s)(2s)^2
Simplifying the expression, we get:
Ksp = 4s^3
Therefore, the Ksp expression of mercury(I) chloride (Hg2Cl2) in terms of molar solubility (s) is 4s^3, considering the stoichiometry of the balanced equation.