Calculate the value of pCl of the following mixed solutions.

1) 50.0ml 0.080M AgNO3 + 50.0ml 0.100M NaCl

2) 40.0ml 0.200M AgNO3 + 80.0ml 0.100M NaCl

3) 25.0ml 0.2M AgNO3 + 50.0ml 0.1M NaCl

4) 20.0ml 0.1M AgNO3 + 50.0ml 0.1M NaCl

(KSP of AgCl = 1.0 × 10^-10)

To calculate the value of pCl for the given mixed solutions, we first need to determine the concentration of chloride ions (Cl^-) and silver ions (Ag⁺) in each solution after mixing. The solubility product constant (K_sp) of AgCl is crucial here, as it helps us understand the conditions under which AgCl will precipitate. Let's break this down step by step for each solution.

1) 50.0 ml 0.080 M AgNO₃ + 50.0 ml 0.100 M NaCl

First, we calculate the moles of Ag⁺ and Cl^- in the mixed solution:

• Moles of Ag⁺ = 0.080 M × 0.050 L = 0.0040 moles
• Moles of Cl^- = 0.100 M × 0.050 L = 0.0050 moles

After mixing, the total volume is 100.0 ml (0.100 L). The concentrations are:

• [Ag⁺] = 0.0040 moles / 0.100 L = 0.040 M
• [Cl^-] = 0.0050 moles / 0.100 L = 0.050 M

Next, we check if AgCl will precipitate by calculating the ion product (Q):

Q = [Ag⁺][Cl^-] = (0.040)(0.050) = 0.0020

Since Q (0.0020) is greater than K_sp (1.0 × 10^-10), AgCl will precipitate. To find the remaining concentration of Cl^- after precipitation, we set up the equilibrium expression:

K_sp = [Ag⁺][Cl^-]
1.0 × 10^-10 = 0 × [Cl^-]

Thus, all Ag⁺ will precipitate, and we can find the remaining Cl^-:

Remaining Cl^- = Initial Cl^- - Ag⁺ = 0.050 M - 0.040 M = 0.010 M

Finally, we calculate pCl:

pCl = -log[Cl^-] = -log(0.010) = 2.00

2) 40.0 ml 0.200 M AgNO₃ + 80.0 ml 0.100 M NaCl

Calculating moles:

• Moles of Ag⁺ = 0.200 M × 0.040 L = 0.0080 moles
• Moles of Cl^- = 0.100 M × 0.080 L = 0.0080 moles

Total volume = 120.0 ml (0.120 L). Concentrations:

• [Ag⁺] = 0.0080 moles / 0.120 L = 0.067 M
• [Cl^-] = 0.0080 moles / 0.120 L = 0.067 M

Calculating Q:

Q = (0.067)(0.067) = 0.004489

Since Q > K_sp, AgCl will precipitate. Remaining Cl^-:

Remaining Cl^- = 0.067 M - 0.067 M = 0 M

Thus, pCl = -log(0) is undefined, indicating complete precipitation.

3) 25.0 ml 0.2 M AgNO₃ + 50.0 ml 0.1 M NaCl

Calculating moles:

• Moles of Ag⁺ = 0.2 M × 0.025 L = 0.0050 moles
• Moles of Cl^- = 0.1 M × 0.050 L = 0.0050 moles

Total volume = 75.0 ml (0.075 L). Concentrations:

• [Ag⁺] = 0.0050 moles / 0.075 L = 0.067 M
• [Cl^-] = 0.0050 moles / 0.075 L = 0.067 M

Calculating Q:

Q = (0.067)(0.067) = 0.004489

Since Q > K_sp, AgCl will precipitate. Remaining Cl^-:

Remaining Cl^- = 0.067 M - 0.067 M = 0 M

Thus, pCl = -log(0) is undefined, indicating complete precipitation.

4) 20.0 ml 0.1 M AgNO₃ + 50.0 ml 0.1 M NaCl

Calculating moles:

• Moles of Ag⁺ = 0.1 M × 0.020 L = 0.0020 moles
• Moles of Cl^- = 0.1 M × 0.050 L = 0.0050 moles

Total volume = 70.0 ml (0.070 L). Concentrations:

• [Ag⁺] = 0.0020 moles / 0.070 L = 0.02857 M