How the solutions were made:

Solution answer solved

Moles of Ag⁺ = Initial moles of Ag⁺ - Moles of KSCN added

Moles of Ag⁺ = 0.014 moles - 1.98×10⁻³ moles

Moles of Ag⁺ in 55 mL = 0.033055 moles

Moles of Ag⁺ that precipitated:

Moles of C2H3O2⁻ in 55 mL solution = Moles of Ag⁺ that precipitated

Moles of C2H3O2⁻ in 55 mL solution = 0.033055 moles

Volume of AgNO3 solution = 35.0 mL = 0.0350 L

Concentration of AgNO3 = 0.250 M

The balanced chemical equation for the reaction between AgNO3 and NaC2H3O2 is:

AgNO3 + NaC2H3O2 → AgC2H3O2 + NaNO3

Volume of NaC2H3O2 solution B = 30.0 mL = 0.0300 L

Concentration of NaC2H3O2 = 0.300 M

To calculate the moles of silver ion (Ag⁺) that formed the silver acetate precipitate and the moles of acetate ion (C2H3O2⁻) in the precipitate, we can use the information provided.

1. Moles of Ag⁺ that formed the silver acetate precipitate: We already have the initial moles of Ag⁺ in solution A, B, and C. Since the reaction between AgNO3 and NaC2H3O2 is a 1:1 ratio, the moles of Ag⁺ are equal to the moles of acetate ion (C2H3O2⁻) in the precipitate.

Therefore, the moles of acetate ion present in the 55.0 mL filtrate is 0.045 moles.