a) Silver iodide(s) is formed by the following reaction:
Ag (aq) + I (aq) --> AgI (s)
The Ksp for AgI = 1.8 x 10^-10
Describe, using the chemical reaction, the mathematical formula for
these ions. Assuming these ion concentrations are equal, what would the molarity be for each ion?
b) Ca3(PO4)2 (s) is a chemical found in bones. If the solution used to preserve these bones contains 10^-5 M Ca^2+ and 10^-6 M PO4(^3-) would the bones soften and dissolve over time? Explain your answer
Ksp = 1.8E-10 = (Ag^+)(I^-)
............AgI(s) ==> (Ag^+)(I^-)
I...........solid........0.....0
C...........solid........x.....x
E...........solid........x.....x
Substitute the E line into the Ksp expression and solve for x.
.......Ca3(PO4)2 ==> 3Ca^2+ + 2PO4^3-
Qsp = (Ca^2+)^3(PO4^3-)^2
Qsp = (10^-5)^3(10^-6)2 = ?
Then compare Qsp with Ksp. Ksp is not listed in the problem.
a) The chemical reaction can be represented as follows:
Ag(aq) + I(aq) → AgI(s)
To determine the mathematical formula for the ions, we can compare the reaction with the chemical formulas of the substances involved. From the reaction, it can be inferred that Ag(aq) represents silver ions (Ag+) and I(aq) represents iodide ions (I-). Similarly, AgI(s) signifies the formation of silver iodide, where Ag+ and I- ions combine to form the solid AgI.
Assuming the ion concentrations are equal, let's denote the molarity of both Ag+ and I- as x.
Therefore, [Ag+] = [I-] = x (since their concentrations are assumed to be equal)
As per the stoichiometry of the reaction, for every 1 mole of AgI formed, 1 mole of Ag+ and 1 mole of I- ions are consumed. Hence, the concentration of AgI in terms of x would be [AgI] = x.
Now, we can write the expression for the solubility product constant (Ksp) using the concentrations of the ions:
Ksp = [Ag+][I-]
= x * x
= x^2
Given that the Ksp for AgI is 1.8 x 10^-10, we can equate it to the expression for Ksp above:
1.8 x 10^-10 = x^2
To find the molarity of each ion, we need to solve this equation for x.
b) To determine if the bones would soften and dissolve over time, we need to compare the solubility product (Ksp) of Ca3(PO4)2 with the product of the concentrations of calcium ions (Ca^2+) and phosphate ions (PO4^3-) present in the solution.
The chemical formula for calcium phosphate (Ca3(PO4)2) indicates that it contains three calcium ions (Ca^2+) and two phosphate ions (PO4^3-) per formula unit.
Given the concentrations of Ca^2+ as 10^-5 M and PO4^3- as 10^-6 M, we can substitute these values into the expression for the ion concentrations:
[Ca^2+] = 10^-5 M
[PO4^3-] = 10^-6 M
Using the stoichiometry, for every 1 mole of calcium phosphate dissolving, 3 moles of calcium ions and 2 moles of phosphate ions are released. Therefore, the concentrations of the ions in terms of x (if all the calcium phosphate dissolves) would be:
[Ca^2+] = 3x
[PO4^3-] = 2x
Comparing this to the given concentrations, we can equate:
3x = 10^-5
2x = 10^-6
Solving these equations, we can find the value of x, which represents the concentration of the ions.
If the concentration of x is lower than the initial concentrations provided, it indicates that the calcium phosphate is not fully dissolved. In this case, the bones would not soften and dissolve over time.