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Identifying Ions In A Mystery Solution, Study Guides, Projects, Research of Chemistry

Fanshawe CollegeChemistry

Identify ions in a mystery solution in Chemistry class

Typology: Study Guides, Projects, Research

2017/2018

Uploaded on 08/21/2018

SkyHendrickson
SkyHendrickson 🇨🇦

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Identifying Ions In A Mystery Solution
Unit One : Performance Task
Sky Hendrickson
Purpose : The purpose of this experiment is to identify matter on a basis of physical and
chemical properties. In this case, using the solubility table.
Question : Which of the following ions are present in the solution :
Barium, Silver, and / or Copper (I)?
Hypotheses : If a precipitation reaction occurs when I mix Sodium Chloride, NaCl (aq),
Sodium Sulphate, Na2SO4 (aq), and/or Sodium Acetate with my mystery solution, then
there are Cu, Ba, and / or Ag ions present. Using the solubility table, I predict that there
will be AgCl, BaSO4, and AgSO4 precipitate. (Silver reacting with Chloride forming a
precipitate.
Experimental Design : I will investigate whether or not the solution contains ions, and
which specific ions are present, (Barium, Silver, and/or Copper) followed by the
procedure outlined in figure 1. Last I’ll confirm the presence of these ions using flame
tests.
Materials : Mystery Solution
Sodium Acetate Solution, NaC2H3O2 (aq)
Sodium Chloride Solution, NaCl (aq)
Sodium Sulfate Solution, NaSO4 (aq)
Safety Glasses
Test Tube Rack
Three Eyedroppers
Centrifuge
pf3
pf4
pf5

Partial preview of the text

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Identifying Ions In A Mystery Solution

Unit One : Performance Task

Sky Hendrickson

Purpose : The purpose of this experiment is to identify matter on a basis of physical and

chemical properties. In this case, using the solubility table.

Question : Which of the following ions are present in the solution :

Barium, Silver, and / or Copper (I)?

Hypotheses : If a precipitation reaction occurs when I mix Sodium Chloride, NaCl (aq),

Sodium Sulphate, Na 2 SO4 (aq), and/or Sodium Acetate with my mystery solution, then

there are Cu, Ba, and / or Ag ions present. Using the solubility table, I predict that there

will be AgCl, BaSO 4 , and AgSO 4 precipitate. (Silver reacting with Chloride forming a

precipitate.

Experimental Design : I will investigate whether or not the solution contains ions, and

which specific ions are present, (Barium, Silver, and/or Copper) followed by the

procedure outlined in figure 1. Last I’ll confirm the presence of these ions using flame

tests.

Materials : Mystery Solution

Sodium Acetate Solution, NaC 2 H 3 O2 (aq)

Sodium Chloride Solution, NaCl (aq)

Sodium Sulfate Solution, NaSO4 (aq)

Safety Glasses

Test Tube Rack

Three Eyedroppers

Centrifuge

Procedure

Step One : Test if there are ions present in the mystery solution, using the conductivity

apparatus.

Step Two : Take your eyedropper #1, containing Sodium Carbonate, NaCO3 (aq), and

transfer 20 to 30 drops into the mystery solution.

Step Three : Observe, if the solution forms a precipitate then it contains Ag+ ions,

precipitated as AgCO 3 (s). If not, there are no Ag+ ions present.

Step Four : If the first mixture formed a precipitate, use the centrifuge to separate the

AgCO 3 (s) from the solution.

Step Five : Filter the mixture using the funnel and separate the AgCO 3 precipitate from

the supernate.

Step Six : Take your eyedropper #2, containing Sodium Chloride, NaCl (aq), and repeat

step one.

Step Seven : Observe, if the solution forms a precipitate then it contains Cu+ ions,

precipitated as CuCl (s). If not, then there are no Cu+ ions present.

Step Eight : If the second mixture formed a precipitate, use the centrifuge to separate

the CuCl (s) precipitate from the solution.

Step Nine : Filter the mixture using a funnel and separate the CuCl precipitate from the

supernate.

Step Ten : Take your eyedropper #3, containing Sodium Sulfate, Na 2 SO 4 and transfer 20

to 30 drops into the mystery solution.

Step Eleven : Observe, if the solution forms a precipitate then it contains Ba+ ions,

precipitated as BaSO 4 (s). If not, there are no Ba+ ions present.

Step Twelve : If the third mixture formed a precipitate, use the centrifuge to separate the

BaSO 4 (s) precipitate from the solution.

BaSO4 (s) Precipitate Light Green

Analysis

I used the conductivity apparatus to find if there were any ions present in the mystery

solution. It showed up positive for ions. The first mixture, contained Ag+ ions, forming a

yellow-ish white precipitate. The second mixture, contained Cu+ ions, forming a white

precipitate. The third mixture, contained Ba2+ ions, forming a white precipitate as well.

Based on the evidence I have gathered in this investigation, the mystery solution that

was prepared for me, contains two of the three, Ag+, and Ba2+ ions.

Evaluation

The design, materials, and skills used in this investigation to identify the ions in the

mystery solution, are acceptable because I used qualitative analysis to support my

answer for the question provided. Using chemical reactions and flame tests made it easy

for me to identify any ions present in the solution, and which specific ions they were,

Cu+, Ag+, and/or Ba2+. The procedure worked very well, and was easy to follow. It was

difficult identifying which ions were present in the first mixture, because CO3 forms a

precipitate with both Ag+, and Ba2+ ions. My hypothesis is supported with the chemical

equations below, which shows that there are Ag+, and Ba2+ ions present in the mystery

solution. Based on both flame tests of the solutions and the precipitates, I was able to

make sure that I was certain that Ag+ ions (showed no colour during the flame test

-solution and precipitate) and Ba2+ ions (showed up a light green -solution and

precipitate) were present in the mystery solution.

Synthesis

I used qualitative analysis by conducting flame tests and precipitation reactions. Both

Silver and Barium formed precipitates. The Barium emit a light green flame during the

flame test, and Silver emit no colour. Flame emission Spectroscopy is used in fireworks,

this is how you can see the different explosions of colour.

Chemical Equations

Cu (aq) + NaC 2 H 3 O2 (aq) →Na + Cu C 2 H 3 O 2 Balanced No Precipitate.

Ba (aq) + NaC 2 H 3 O2 (aq) → Na + Ba(C 2 H 3 O 2 ) 2

Ba (aq) + 2 NaC 2 H 3 O2 (aq) → 2Na + Ba(C 2 H 3 O 2 ) 2 No precipitate.

Ag(aq) + NaC 2 H 3 O 2 → Na (aq) + AgC 2 H 3 O2 (aq) Balanced No Precipitate.

___________________________________________________________

Chemical Changed

Cu (aq) + Na 2 CO3 (aq) → Na (aq) + Cu 2 CO3 (aq) Solubility Data Not Available.

2Cu (aq) + Na 2 CO3 (aq) → 2 Na (aq) + Cu 2 CO3 (aq)

Ba (aq) + Na 2 CO3 (aq) → Na (aq) + Ba 2 CO3 (aq)

2Ba (aq) + Na 2 CO3 (aq) → 2Na (aq) + Ba 2 CO3 (s) Ba 2 CO3 (s) precipitate is formed.

Ag (aq) + Na 2 CO 3 →Na (aq) + Ag 2 CO 3

2Ag (aq) + Na 2 CO 3 → Na (aq) + Ag 2 CO 3 Ag 2 CO3 (s) precipitate is formed.

__________________________________________________________

Cu (aq) + NaCl (aq) → Na (aq) + CuCl (aq) Balanced No Precipitate.

Ba (aq) + NaCl (aq) → Na (aq) + BaCl2 (aq)

Ba (aq) + 2 NaCl (aq) → 2 Na (aq) + BaCl2 (aq) No Precipitate.

Ag (aq) + NaCl (aq) → Na (aq) + AgCl (s) Balanced AgCl precipitate is formed.

___________________________________________________________

Cu (aq) + Na 2 SO4 (aq) →Na (aq) + Cu 2 SO4 (aq)

2Cu (aq) + Na2SO4 (aq) → 2Na (aq) + Cu2SO4 (aq) No Precipitate.

Ba (aq) + Na 2 SO4 (aq) → Na (aq) + BaSO4 (s)

Ba (aq) + Na 2 SO4 (aq) → 2 Na (aq) + BaSO4 (s) Forms BaSO4 Precipitate.