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

Step by step solution

01

(Step 1: Moles of Ag+ in the solution)

(First, we need to find the moles of Ag+ in the 15.0 mL of 0.200 M AgNO3 solution. We can do this using the formula: Moles of Ag+ = Molarity of AgNO3 * Volume of AgNO3 solution Moles of Ag+ = 0.200 mol/L * 15.0 mL * (1 L / 1000 mL) Moles of Ag+ = 0.003 mol)

02

(Step 2: Moles of HCl needed to precipitate Ag+)

(Next, we need to find the moles of HCl needed to precipitate all the silver ions. The balanced chemical equation for the reaction is: AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq) From the balanced equation, we can see that 1 mole of HCl is required for each mole of AgNO3. Moles of HCl needed = Moles of Ag+ Moles of HCl needed = 0.003 mol)

03

(Step 3: Volume of 0.150 M HCl required)

(Now we can find the volume of the 0.150 M HCl solution needed to precipitate the silver ions: Volume of HCl = Moles of HCl / Molarity of HCl Volume of HCl = 0.003 mol / 0.150 mol/L Volume of HCl = 0.02 L or 20 mL)

04

(Step 4: Moles of KCl needed to precipitate Ag+)

(The balanced chemical equation for the reaction of AgNO3 with KCl is: AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq) From the balanced equation, we can see that 1 mole of KCl is required for each mole of AgNO3. Moles of KCl needed = Moles of Ag+ Moles of KCl needed = 0.003 mol)

05

(Step 5: Mass of KCl needed)

(Now we will find the mass of KCl needed to precipitate the silver ions: Mass of KCl = Moles of KCl * Molar mass of KCl Mass of KCl = 0.003 mol * 74.55 g/mol Mass of KCl = 0.224 g)

06

(Step 6: Cost comparison)

(First, we calculate the cost of using 0.150 M HCl solution: Cost of HCl = (Volume of HCl used / Total volume of HCl) * Cost of HCl solution Cost of HCl = (20 mL / 500 mL) * $39.95 Cost of HCl = $1.598 Next, we calculate the cost of using KCl: Cost of KCl = (Mass of KCl used / Total mass of KCl) * Cost of KCl Cost of KCl = (0.224 g / 1,000,000 g) * $10 Cost of KCl = $0.00000224 Now, we can compare the costs: Cost for HCl: $1.598 Cost for KCl: $0.00000224 Clearly, using KCl is more cost-effective for this analysis.)

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