Question

1. An experiment calls for you to use 100 mL of 0.25 M HNO3 solution. All you have available is a bottle of 3.4 M HNO3. How many milliliters of the 3.4 M HNO3 solution do you need to prepare the desired solution?

2. How many milliliters of water do you need to prepare the desired solution?

A. You could add HCl(aq) to the solution to precipitate out AgCl(s). What volume of a 0.160 M HCl(aq) solution is needed to precipitate the silver ions from 12.0 mL of a 0.160 M AgNO3 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 12.0 mL of 0.160 M AgNO3 solution?

C. Given that a 0.160 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

1) M1V1 = M2V2

3.4 M * V1 = 0.25 M * 100 mL

V1 = 7.35 ml

2) HCl + AgNO3 → AgCl + HNO3

(12.0 mL) x (0.160 M AgNO3) x (1 mol HCl / 1 mol AgNO3) / (0.160 M HCl) = 12 mL HCl

b)
KCl + AgNO3 → AgCl + KNO3 [balanced as written]

(0.0120 L) x (0.160 mol/L AgNO3) x (1 mol KCl / 1 mol AgNO3) x (74.5513 g KCl/mol) = 0.143 g KCl

c) HCl is more cost-effective by far

as 500ml = 39.95 $

1ml = 39.95 $ / 500

12 ml = 39.95 $ *12/ 500 = 0.95 $ for 12 m l of HCL

1ton = 907185 g = 10 $

1 g = 10 / 907185

0.143 = 10* 0.143 / 907185 = 1.57 * 10 ^-7 $