You want to analyze a silver nitrate solution. (a) You could add \(\mathrm{HCl}(a q)\) to the solution to precipitate out \(\mathrm{AgCl}(s) .\) What volume of a \(0.150 \mathrm{MHCl}(a q)\) solution is needed to precipitate the silver ions from \(15.0 \mathrm{~mL}\) of a \(0.200 \mathrm{MAgNO}_{3}\) solution? (b) 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 \(15.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) solution? (c) Given that a $0.150 \mathrm{MHCl}(a q)\( solution costs \)\$ 39.95\( for \)500 \mathrm{~mL}$ and that KCl costs \(\$ 10 /\) ton, which analysis procedure is more cost-effective?

Step by step solution

01

Calculate moles of AgNO3

First, we will find the moles of AgNO3 present in the 15.0 mL of 0.200 M solution. We can calculate it using the following formula: Number of moles = Molarity × Volume //Convert volume to liters: \(15.0\:mL = 0.015\:L\) Moles of AgNO3 = \(0.200\:\cancel{Mol\:L^{-1}}\cdot0.015\:\cancel{L} = 0.003\:mol\)

02

Calculate moles of HCl needed

Since one mole of AgNO3 reacts with one mole of HCl to produce AgCl, the number of moles of HCl required is equal to the moles of AgNO3. So, we need 0.003 moles of HCl.

03

Calculate volume of 0.150 M HCl solution needed

Now, we will use the number of moles of HCl and its molarity to find the volume of the solution needed: Volume = Moles / Molarity Volume = \(0.003\:\cancel{mol}/0.150\:\cancel{Mol\:L^{-1}} = 0.02\:L\) Volume = \(0.02\:L = 20\:mL\) So, we need 20 mL of 0.150 M HCl solution to precipitate all the silver ions. #b) Calculating the mass of KCl needed#

04

Calculate moles of KCl needed

Again, one mole of AgNO3 reacts with one mole of KCl to produce AgCl, so we need 0.003 moles of KCl.

05

Calculate mass of KCl needed

Now, we'll use the molar mass of KCl to find the mass of KCl required: Mass of KCl = Moles × Molar mass Mass of KCl = \(0.003\:mol \times 74.55\:g/mol = 0.22365\:g\) So, we need 0.22365 g of KCl to precipitate all the silver ions. #c) Comparing costs of the two procedures#

06

Calculate the cost of HCl solution needed

To calculate the cost of the HCl solution, we need to find the cost of 20 mL of 0.150 M HCl solution. We are given that 500 mL of the solution costs $39.95, so the cost per milliliter is: Cost per mL = Total cost / Volume Cost per mL = \(39.95 / 500 mL = \)0.0799/mL Now, we can find the cost of 20 mL: Cost of 20 mL = 20 mL × \(0.0799/mL = \)1.598

07

Calculate the cost of KCl needed

In this case, we need to find the cost of 0.22365 g of KCl. We are given that 1 ton (= 1,000,000 g) of KCl costs $10, so the cost per gram is: Cost per g = Total cost / Mass Cost per g = \(10/ 1,000,000 g = \)0.00001/g Now, we can find the cost of 0.22365 g: Cost of 0.22365 g = \(0.00001/g × 0.22365 g = \)0.0000022365 ≈ $0.000002

08

Compare the costs

Now that we have the costs of both procedures: Cost of HCl solution: ≈ $1.598 Cost of KCl: ≈ $0.000002 The cost of using KCl is significantly lower than using HCl solution. Therefore, adding solid KCl to the solution to precipitate the silver ions is the more cost-effective procedure.

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