The solubility products of AgCl and 𝐴𝑔3𝑃𝑂4 are 1.6 × 10−10 and 1.8 × 10−18, respectively. If 𝐴𝑔+ is added (without changing the volume) to 1.00 L of a solution containing 0.10 mol 𝐶𝑙− and 0.10 mol 𝑃𝑂4 3−, calculate the concentration of 𝐴𝑔+ions (in mol/L) required to initiate

Question

Create an illustrative, non-textual image for a chemistry-themed question. Depict a transparent glass beaker containing a clear solution against the backdrop of a classic white laboratory environment. Show silver ions being added to the solution in the beaker, symbolized by tiny, shimmering, silver particles. The particles interact with existing elements in the solution, demonstrating potential reaction to form silver chloride and silver phosphate, visually presented as distinct, forming crystals within the solution. Ensure to achieve a soft, realistic scientific illustration style.

The solubility products of AgCl and 𝐴𝑔3𝑃𝑂4 are 1.6 × 10−10 and 1.8 × 10−18, respectively. If 𝐴𝑔+ is added (without changing the volume) to 1.00 L of a solution containing 0.10 mol 𝐶𝑙− and 0.10 mol 𝑃𝑂4 3−, calculate the concentration of 𝐴𝑔+ions (in mol/L) required to initiate

(a) the precipitation of AgCl and
(b) the precipitation of 𝐴𝑔3𝑃𝑂4

Answer 1

To calculate the concentration of Ag+ ions required to precipitate AgCl and Ag3PO4, we can use the solubility product (Ksp) equation for each compound.

(a) The solubility product (Ksp) equation for AgCl is given as:

Ksp = [Ag+][Cl-]

Let's break down the problem step by step:

Step 1: Write the balanced equation for the dissociation of AgCl:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

Step 2: Calculate the initial concentration of Ag+ and Cl- ions:
Since we are given that the solution contains 0.10 mol Cl- ions, the initial concentration of Cl- ions is 0.10 mol/L.
The initial concentration of Ag+ ions is zero since no Ag+ ions are initially present.

Step 3: Let's assume that 'x' mol/L of Ag+ ions are required to reach equilibrium and precipitate AgCl.

At equilibrium:
[Ag+] = x mol/L
[Cl-] = 0.10 mol/L

Step 4: Substitute the concentrations into the Ksp equation and solve for x:
Ksp = [Ag+][Cl-]
1.6 × 10^(-10) = (x)(0.10)

Solve the equation for x:
x = 1.6 × 10^(-10) / 0.10

x ≈ 1.6 × 10^(-9) mol/L

Therefore, the concentration of Ag+ ions required to initiate the precipitation of AgCl is approximately 1.6 × 10^(-9) mol/L.

(b) The solubility product (Ksp) equation for Ag3PO4 is given as:

Ksp = [Ag+]^3[PO4^3-]

Let's break down the problem step by step:

Step 1: Write the balanced equation for the dissociation of Ag3PO4:
Ag3PO4(s) ⇌ 3Ag+(aq) + PO4^3-(aq)

Step 2: Calculate the initial concentration of Ag+ and PO4^3- ions:
Since we are given that the solution contains 0.10 mol PO4^3- ions, the initial concentration of PO4^3- ions is 0.10 mol/L.
The initial concentration of Ag+ ions is zero since no Ag+ ions are initially present.

Step 3: Let's assume that 'y' mol/L of Ag+ ions are required to reach equilibrium and precipitate Ag3PO4.

At equilibrium:
[Ag+] = 3y mol/L
[PO4^3-] = 0.10 mol/L

Step 4: Substitute the concentrations into the Ksp equation and solve for y:
Ksp = [Ag+]^3[PO4^3-]
1.8 × 10^(-18) = (3y)^3(0.10)

Solve the equation for y:
y = ∛(1.8 × 10^(-18) / (0.10 * 3^3))

y ≈ 7.60 × 10^(-7) mol/L

Therefore, the concentration of Ag+ ions required to initiate the precipitation of Ag3PO4 is approximately 7.60 × 10^(-7) mol/L.

Answer 2

To calculate the concentration of Ag+ ions required for the precipitation of AgCl and Ag3PO4, we need to compare the solubility products (Ksp) of the two compounds with the initial concentrations of Cl- and PO4^3- ions in the solution.

(a) Precipitation of AgCl:
The balanced equation for the precipitation of AgCl is:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

The Ksp expression for AgCl is:
Ksp = [Ag+][Cl-]

Given that the Ksp of AgCl is 1.6 × 10^(-10) and the initial concentration of Cl- ions is 0.10 mol/L, we can set up the equation as follows:
1.6 × 10^(-10) = [Ag+][0.10]

Solving for [Ag+]:
[Ag+] = (1.6 × 10^(-10))/(0.10)

Calculating the concentration of Ag+ required to initiate the precipitation of AgCl will give you the answer.

(b) Precipitation of Ag3PO4:
The balanced equation for the precipitation of Ag3PO4 is:
3AgPO4(s) ⇌ Ag3PO4(s) + 3Ag+(aq) + PO4^3-(aq)

The Ksp expression for Ag3PO4 is:
Ksp = [Ag+]^3[PO4^3-]

Given that the Ksp of Ag3PO4 is 1.8 × 10^(-18) and the initial concentrations of PO4^3- ions is 0.10 mol/L, we can set up the equation as follows:
1.8 × 10^(-18) = [Ag+]^3[0.10]

Solving for [Ag+]:
[Ag+] = (1.8 × 10^(-18))^(1/3) / (0.10)

Calculating the concentration of Ag+ required to initiate the precipitation of Ag3PO4 will give you the answer.

Note: It is important to consider the common ion effect when calculating the solubility of compounds in the presence of common ions. The presence of Ag+ ions can reduce the solubilities of AgCl and Ag3PO4.

Answer 3

What (Ag^+) is required to ppt AgCl and Ag3PO4 in separate solutions?

For AgCl it is (Ag^+) = (1.6E-10/0.1) = 1.6E-9 M
For Ag3PO4 it is (Ag^+) = (1.8E-18/0.1)^1/3 = 2.6E-6 M
(a) Therefore, Ag^+ added drop wise to the initial solution will ppt AgCl first when (Ag^+) = 1.6E-9 M and leave Ag3PO4 undisturbed.

(b) When (Ag^+) = 2.62E-6 M the Ag3PO4 will begin ppting.

Well done

To calculate the concentration of Ag+ ions required to precipitate AgCl and Ag3PO4, we can use the solubility product (Ksp) equation for each compound.

To calculate the concentration of Ag+ ions required for the precipitation of AgCl and Ag3PO4, we need to compare the solubility products (Ksp) of the two compounds with the initial concentrations of Cl- and PO4^3- ions in the solution.

What (Ag^+) is required to ppt AgCl and Ag3PO4 in separate solutions?