In a saturated solution of silver phosphate, the concentration of silver ion is 4.5 x 10^-4 mol/L. The Ksp of silver phosphate would be which of the following?
6.8 10-8 1.0 10-11
1.4 10-14 none of the above
1.5 10-15
In a saturated solution of aluminum hydroxide, the concentration of aluminum ion is 2.4 x 10^-8 mol/L. The Ksp of aluminum hydroxide would be?
In a saturated solution of PbBr2, the concentration of Br1- is 8.2 x 10^-2 mol/L. The [Pb2+] in mol/L is?
here on june 18, 2021
n a saturated solution of silver phosphate, the concentration of silver ion is 4.5 x 10^-4 mol/L. The Ksp of silver phosphate would be which of the following?
Ag3PO4 ==> 3Ag^+ + PO4^-3
Ksp = (Ag^+)^3(PO4^-3)
(Ag^+) from the problem is 4.5 x 10^-4.
(PO4^-3) = 1/3 of that.
Substitute into Ksp and solve.
In a saturated solution of aluminum hydroxide, the concentration of aluminum ion is 2.4 x 10^-8 mol/L. The Ksp of aluminum hydroxide would be?
Same process as Ag3PO4.
In a saturated solution of PbBr2, the concentration of Br1- is 8.2 x 10^-2 mol/L. The [Pb2+] in mol/L is?
PbBr2 ==> Pb^+2 + 2Br^-
Ksp = (Pb^+2)(Br^-)^2
Shouldn't (Pb^+2) = 1/2 the Br^-?
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Why did the solution go to therapy?
It had some serious concentration issues! Now let's calculate those Ksp values and [Pb2+] for you.
For the first question, the Ksp of silver phosphate would be 1.5 x 10^-15. I hope it's not too much silver for your taste!
For the second question, the Ksp of aluminum hydroxide would be none of the above. Looks like it's keeping its Ksp a secret!
Now, for the third question, the concentration of Br1- in PbBr2 is 8.2 x 10^-2 mol/L. We'll need to do some math (brace yourself):
PbBr2 dissociates into Pb2+ and 2Br1-. So if the concentration of Br1- is 8.2 x 10^-2 mol/L, then the concentration of Pb2+ would be half of that. That means [Pb2+] = 4.1 x 10^-2 mol/L. Phew, we made it! Just don't go spreading that lead around too much.
Brilliant questions, by the way!
To solve these problems, we need to understand the concept of solubility product constant (Ksp). Ksp is an equilibrium constant that represents the maximum product of the concentrations of the ions in a saturated solution at a given temperature. It is used to determine the solubility of a slightly soluble compound in water.
To find Ksp, we need to use the concentrations of the ions in the saturated solution. The general formula for a solubility equilibrium is:
AxBy (s) ⇌ xAy+(aq) + yBx-(aq)
Where A represents the cation, B represents the anion, and x and y represent the stoichiometric coefficients.
For the first problem, the solubility equilibrium is:
Ag3PO4 (s) ⇌ 3Ag+(aq) + PO4^3-(aq)
The concentration of silver ion is given as 4.5 x 10^-4 mol/L. We know that 1 mole of Ag3PO4 produces 3 moles of Ag+ ions, so the concentration of Ag3PO4 is (4.5 x 10^-4 mol/L) / 3 = 1.5 x 10^-4 mol/L. We can use these concentrations to calculate Ksp:
Ksp = [Ag+]^3 [PO4^3-]
= (4.5 x 10^-4)^3 (1.5 x 10^-4)
= 9.1125 x 10^-17
Comparing this value with the given options, we can see that 9.1125 x 10^-17 is closest to 1.4 x 10^-14. Therefore, the correct answer is 1.4 x 10^-14.
For the second problem, the solubility equilibrium is:
Al(OH)3 (s) ⇌ Al^3+(aq) + 3OH-(aq)
The concentration of aluminum ion is given as 2.4 x 10^-8 mol/L. We can use this concentration to calculate Ksp:
Ksp = [Al^3+] [OH-]^3
= (2.4 x 10^-8) (2.4 x 10^-8)^3
= 13.824 x 10^-32
This value is not listed in the given options, so the correct answer is "none of the above."
For the third problem, the solubility equilibrium is:
PbBr2 (s) ⇌ Pb^2+(aq) + 2Br-(aq)
The concentration of Br^- ion is given as 8.2 x 10^-2 mol/L. We know that 1 mole of PbBr2 produces 2 moles of Br^- ions, so the concentration of PbBr2 is (8.2 x 10^-2 mol/L) / 2 = 4.1 x 10^-2 mol/L. Hence, the concentration of Pb^2+ is equal to the concentration of PbBr2, which is 4.1 x 10^-2 mol/L.
Therefore, the [Pb^2+] in mol/L is 4.1 x 10^-2.
Remember that calculating the solubility product constant (Ksp) requires understanding the stoichiometry of the balanced equation and the concentrations of the ions in the solution.