A 25.00 mL sample of 0.1306 M AgNO3 is being titrated with 0.1194 M KSCN. Calculate the pAg after the addition of 30.0 mL KSCN. Ksp for AgSCN= 1.1x10 12 A) 8.12 B) 9.72 C) 10.39 D) 12.1

A 25.00-ml sample of 0.1306 m AgNO3 is being titrated with 0.1194 m KSCN. Calculate the pag after the addition of 30.0 ml KSCN. Ksp for AgSCN= 1.1x10-12 a) 8.12 b) 9.72 c) 10.39 d) 12.1

Accepted Answer

Here's how to solve the problem:
1. Determine the moles of Ag+ initially present:

Moles of Ag+ = (Volume of AgNO3 solution)
(Concentration of AgNO3)
Moles of Ag+ = (25.00 mL)
(0.1306 mol/L) = 3.265 mmol
2. Determine the moles of SCN- added:

Moles of SCN- = (Volume of KSCN solution)
(Concentration of KSCN)
Moles of SCN- = (30.00 mL)
(0.1194 mol/L) = 3.582 mmol
3. Calculate the moles of Ag+ remaining:

Since Ag+ and SCN- react in a 1:1 ratio, the limiting reagent is Ag+. The excess SCN- will be present in solution.
Moles of Ag+ remaining = Initial moles of Ag+ - moles of SCN- added
Moles of Ag+ remaining = 3.265 mmol - 3.582 mmol = -0.317 mmol
4. Calculate the concentration of Ag+ remaining:

Concentration of Ag+ = (Moles of Ag+ remaining) / (Total volume of the solution)
Concentration of Ag+ = (-0.317 mmol) / (25.00 mL + 30.00 mL) = -0.00594 mol/L
5. Calculate pAg:

pAg = -log[Ag+]
pAg = -log(-0.00594) = 2.23
6. Consider the effect of the solubility product (Ksp):

Since the concentration of Ag+ is negative, it indicates that Ag+ has been completely consumed, and the solution is saturated with AgSCN. In this case, we need to use the Ksp to calculate the concentration of Ag+.

Ksp = [Ag+][SCN-]
1.1 x 10^-12 = [Ag+][SCN-]
Since [Ag+] = [SCN-], we can write:
1.1 x 10^-12 = [Ag+]^2
[Ag+] = √(1.1 x 10^-12) = 1.05 x 10^-6 M

pAg = -log[Ag+]
pAg = -log(1.05 x 10^-6) = 5.98
Therefore, the correct answer is closest to 5.98. None of the provided options are close to this value.

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