(Need the solutions to these problems, not the answers) 1. A solution whose conc

Question

(Need the solutions to these problems, not the answers)

1. A solution whose concentration of Na+ is 11.0 ppm is prepared by dissolving sodium sulfate in water. Calculate:

a) Analytical (molar) concentration of the solution

b) Species or equilibrium (molar) concentration

2. You are asked to prepare 500.0 mL of each of the following solutions from the reagent indicated. Calculate how much reagent you would need to weigh out in each case.

a) 7.0 parts per thousand K+ from potassium chloride

b) 1.0x102 ppb Zn2+ from zinc nitrate

c) 50.0 ppm Br- from sodium bromide

3. 10.63 g of copper(II) nitrate are dissolved in water to produce 500.0 mL of solution. Calculate:

a) The formal molar concentration of the solution

b) The analytical molar concentration of copper(II) ions

c) The analytical molar concentration of nitrate ions

Answers:

1. a) 2.39x10-4 MNa2SO4
b) 4.78x10-4M Na, 2.39x10-4SO42-

2. a) 6.7g KCl
b) 1.4x10-4g Zn(NO3)2
c) 3.22x10-2g NaBr

3. a) 0.1134 M
b) 0.1134 M
c) 0.2267 M

Explanation / Answer

11g / 1000L = (11/23) mol / 1000 L = 4.7826e-4 M Na+

Na2SO4 -> 2Na+ + SO42-

So, if Na+ = 4.78e-4 M, SO42- = 2.39e-4, Na2SO4 = 2.39e-4

-----------

7 ppk K+ = 7/39/1 M = 1.7949e-1 M

KCl --> K+ + Cl-

So, KCL, Cl- will also be 1.7949e-1 M

For 500 ml, 1.7949e-1 / 2 mol = 8.9744e-2 mol = (39+35.5)*8.9744e-2

= 6.686 g = 6.7 g

--------------

1 ppb = 1 e-9

10^2 ppb = 0.10 ppm

Zn(NO3)2---> Zn2+ + 2NO3-

(0.10/65.4) /1000 M Zn2+ = 1.53e-6 M Zn2+

Zn(NO3)2 = 1.53e-6 M x 0.5 L = 0.765 e-6 moles = 0.765e-6 * (65.4 + 2x14 + 2x3x16)

= 1.4e-4 g

---------------------

50.0 ppm Br= from NaBr

(50.0/80)/1000 M = 6.25e-4 M

NaBr ---> Na+ + Br-

So, NaBr = 0.5*6.25e-4*(23 + 80) g = 3.22e-2 g

--------------------

10.63 g in 500 mL of

Cu(NO3)2 ---> Cu2+ + 2NO3-

So, 10.63 / 0.500 / (63.54 + 2x14 + 2x3x16) M = 0.1134 M solution

Cu(NO3)2 = 0.1134 mol

Cu2+ = 0.1134 mol

NO3- = 2 x 0.1134 mol = 0.2267 mol

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