You could add HCl(aq) to the solution to precipitate out AgCl(s). What volume of a 0.200 M HCl(aq) solution is needed to precipitate the silver ions from 14.0 mL of a 0.180 M AgNO3 solution?

Question

Part 2:

You could add solid KCl to the solution to precipitate out AgCl(s). What mass of KCl is needed to precipitate the silver ions from 14.0 mL of 0.180 M AgNO3 solution?

Part 3:

Given that a 0.200 M HCl(aq) solution costs $39.95 for 500 mL, and that KCl costs $10/ton, which analysis procedure is more cost-effective?

Answer 1

The first question is simple stoichoimetry problem



What is the balanced reaction – AgNO_3(aq) + HCl_(aq) −> AgCl_(s) + HNO_3(aq)

How many moles of  AgNO_3(aq) are present?

14.0 mL * 1 L/1000 mL * .180 M = 2.52 x 10^-3 moles AgNO_3(aq)

Since there is 1 mole of Ag in 1 mole of AgNO_3(aq), there are . 2.52 x 10^‑3 moles Ag present.

This will require 2.52 x 10^‑3 moles HCl (because of the 1:1 ratio)

The volume of HCl would be 2.52 x 10^‑3 moles/(.200 mol/L) * 1000 mL/L = 12.6 mL

Part 2:        You could add solid KCl to the solution to precipitate out AgCl(s). What mass of KCl is needed to precipitate the silver ions from 14.0 mL of 0.180 M AgNO₃ solution?

Again, we start with the balanced reaction; note the solid KCl would dissolve in the solution.

AgNO_3(aq) + KCl_(aq) à AgCl_(s) + KNO_3(aq)

The moles of AgNO_3(aq) would be the same as before

14.0 mL * 1 L/1000 mL * .180 M = 2.52 x 10^-3 moles AgNO_3(aq)

The number of moles of KCl would be equal to the moles of AgNO_3(aq) = 2.52 x 10^-3 moles KCl.

The formula mass for KCl is 74.55 g/mol, so there are 2.52 x 10^-3 moles KCl x 74.55 g/mol = .188 grams of KCl needed to precipitate the silver.

Part 3:        Given that a 0.200 M HCl(aq) solution costs $39.95 for 500 mL, and that KCl costs $10/ton, which analysis procedure is more cost-effective?

This is an interesting question.

Based on the amounts calculated, it would cost $0.99 (12.6 mL * $39.95/500 mL) for the HCl(aq) 

It would cost $1.00 x 10^-5 for the KCl. This is, of course, a relatively meaningless number.

The question arises can you buy limited amounts of the reagents at the prices quoted or must you buy the amount listed. Second question, what will you do with 1 ton of leftover KCl?

The most efficient procedure would be to use the HCl solution, if for no other reason it is easier to store the leftover solution.

Actually, it would be far more efficient to buy the HCl as a stock solution and make the necessary solution.