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Qualitative Analysis:

Quantitative Analysis:

An analysis that determines what’s in a solution, the qualities of the solution.

An analysis that determines the amounts of a substance present in a solution.

Qualitative AnalysisExamples:

1. Solution Colour

Colorimetry:An analysis that uses the colour of a solution and the light that passes through it to analyze the solution.

Qualitative AnalysisExamples:

1. Solution Colour

2. Flame Test

Data Booklet Pg. 6

Quantitative AnalysisExamples:

1. Gravimetric Analysis

The process of using stoichiometry to calculate masses of unknown substances.

Gravimetric (solids)Stoichiometry

Gas StoichiometrySolution Stoichiometry

Quantitative AnalysisExamples:

1. Gravimetric Analysis

2. Titration AnalysisAn analysis that records known volumes and the principles of stoichiometry to calculate concentration, volume, or moles of unknown substance.

***Know your solution stoichiometry really well to do these***

In gravimetric analysis you are using masses and converting them to moles so that you can do stoichiometry.

In gravimetric analysis it is often beneficial to mix two solutions together and make a solid precipitate. (solids are easier to weigh and measure).

Solid precipitates are produced when ions come together and they form slightly soluble compounds that “fall out of solution”.

Example:

AgNO3 + NaCl NaNO3 + AgCl

Predict the products and balance.

1 1 1 1

In gravimetric analysis you are using masses and converting them to moles so that you can do stoichiometry.

In gravimetric analysis it is often beneficial to mix two solutions together and make a solid precipitate. (solids are easier to weigh and measure).

Solid precipitates are produced when ions come together and they form slightly soluble compounds that “fall out of solution”.

Example:

AgNO3 + NaCl NaNO3 + AgCl1 1 1 1

Now use the solubility chart to determine the states…soluble (aq) or slightly soluble (s)

(aq) (aq) (aq) (s)

Precipitate

AgNO3 + NaCl NaNO3 + AgCl1 1 1 1(aq) (aq) (aq) (s)

Precipitate

When your mixing two solutions together you have no idea what the concentration of the two solutions might be.

In this situation you want to produce as much solid so you can find out the moles and thus the concentration, but in order to do this you have to make sure ALL the limiting reagent reacts and produces solid product.

The extent to which ALL the limiting reagent is reacted by adding more and more excess reagent in a step wise fashion is called Precipitation Completeness.

****You continue to add more and more excess reagent in a drop wise fashion until the mixing of the two liquids NO LONGER produces any “cloudiness”.

In any chemical reaction there is one reagent that will “run out” first….limiting reagent.

And one reagent that is present in more amount than can ever be reacted….excess reagent.

The question is, how can you tell?

Whenever you are asked to calculate excess and limiting reagents you will know the amounts of BOTH reactants.

Example

1. Balanced Chemical Equation

Cu + AgNO3 Cu(NO3)2 + Ag

States?

(s) (aq)1 12 2

(s)(aq)

Example

2. Write down what you know.

Cu + AgNO3 Cu(NO3)2 + Ag(s) (aq)1 12 2

(s)(aq)m = 10.0 g m = 20.0 g

Example

3. Convert all quantities to mols.


(s)(aq)m = 10.0 g m = 20.0 gM = 63.55 g/mol

n = 0.157mol

M = 169.88 g/moln = 0.118 mol

Example

4. Convert one reagent into the other using N/G and see which is larger/bigger.


(s)(aq)m = 10.0 g m = 20.0 gM = 63.55 g/mol

n = 0.157mol

M = 169.88 g/moln = 0.118 mol

( )NG21 (0.157 mols)(2/1) = 0.317 mol

Because more mols of AgNO3 can be produced than are actually present, AgNO3 is the limiting reagent.

L.RE.R

Example


(s)(aq)m = 10.0 g m = 20.0 gM = 63.55 g/mol

n = 0.157mol

M = 169.88 g/mol

n = 0.118 mol

L.RE.R

Once you have determined the L.R and E.R, you can used the limiting reagent amount to calculate how much MAX Silver(Ag) will be produced.

( )22

n = 0.118 mol

m = ?m = 12.7 g

Example


(s)(aq)m = 10.0 g m = 20.0 gM = 63.55 g/mol

n = 0.157mol

M = 169.88 g/mol

L.RE.R

You can also use the amount of the L.R to determine how much of the E.R will be used up….subtract these two amounts and you can calculate how much is left over.

n = 0.118 mol

m = 12.7 g

( )12

n = 0.0589 mol

nCu(s) Left Over= Mols of E.R – mols from L.R= 0.157mol – 0.0589mol= 0.0981

n = 0.0981

Example


(s)(aq)m = 10.0 g m = 20.0 gM = 63.55 g/mol

n = 0.157mol

M = 169.88 g/mol

L.RE.R

You can calculate the MASS of Cu left over by converting the mols into grams by multiplying by the molar mass.

n = 0.118 mol

m = 12.7 g

n = 0.0981 Mass Cu(s) = nM = (0.0981 mol)(63.55 g/mol) = 6.23 g

Example 2

FeCl3(aq) + NaOH(aq) Fe(OH)3(aq) + NaCl(aq)1 33 1m = 26.8 g m = 21.5 g

n = m M

n = 0.165 mol n = 0.538 mol

( )31

n = 0.496 mol

L.R E.R

0.538mol – 0.496 mol = mols of NaOH left over.

0.042 mol NaOH left over.

Example 2


n = 0.165 mol n = 0.538 mol

L.R E.R

0.042 mol NaOH left over.n = m M

m = nMm = (0.042 mol)(40.00 g/mol) = 1.7 g of NaOH left over.

Example 2


n = 0.165 mol

n = 0.538 mol

L.R E.R 1.7 g of NaOH left over.

****To determine the mass of each product you use the L.R and do stoichiometry.****

m = ? m = ?

( )11

n = 0.165 mol

m = nM

m = 17.7 g

Example 2


n = 0.165 mol

n = 0.538 mol

L.R E.R 1.7 g of NaOH left over.

****To determine the mass of each product you use the L.R and do stoichiometry.****

m = ?

( )31

m = nM

m = 17.7 g

n = 0.495 mol

m = 29.0 gDONE

Titration is an experiment where you KNOW the concentrations AND volume of one solution (mols) and you use it determine the concentration of the other solution.

You set up the calculations exactly the same as a regular stoichiometry.

Titrant:

Sample:

Is the solution you put in the burette (long glass thing).

Is the solution in the beaker.

In titration you know enough information about either the titrant or the sample to determine how many moles are present using n = CV.

During any titration as you are adding drops from the burette (mols), there will eventually be a moment when the mols of the two solutions are equal, called the “equivalence point”.

But we cant see mols, so we need some way to determine when we have equal mols so we can stop and do calculations.

Different solutions each have different prosperities, like pH.

So we put a pH indicator in the sample so that it will change colour when we have equal mols, telling us to “end” and the titration.

This colour change is called the “End Point”.

Lets Try One….It Will Make More Sense After You Try One For Real.

Step 1: Balanced Chemical Equation.

Na2CO3(aq) + HCl(aq) H2CO3(aq) + NaCl(aq)1 22 1

Na2CO3(aq) + HCl(aq) H2CO3(aq) + NaCl(aq)1 22 1Step 2: Write down what you know.

n = m M

V = 0.100 LC = ?m = 1.59 g

n = 1.59 g 105.99 g/mol

n = 0.01500142 mol

n = 0.0150 mol


V = 0.100 LC = ?n = 0.0150 mol

C = n V

= 0.0150 mol 0.100L

C = 0.150 mol/L

C = 0.150 mol/L


C = ?C = 0.150 mol/LV = 0.010 L

The information above was collected after completing 4 trials of titrations.

You do titrations multiple times to make sure you didn’t mix too much together and add more than you needed to.

You want to take the THREE closes trials and average them….scratch the trial that is way off.

12.7 + 12.8 + 12.8 3 trials

= 12.8 ml This is the volume of HCl needed to get equal mols.


C = ?C = 0.150 mol/LV = 0.010 L V = 0.0128 L

From now on its just solution stoichiometry.

n = CV

n = 0.0015 mol( )21

n = 0.003 mol


C = ?C = 0.150 mol/LV = 0.010 L V = 0.0128 L

n = 0.003 mol

C = 0.003 mol 0.0128 L

C = n V

C = 0.234 mol/L

C = 0.234 mol/LDONE

During a titration you are either adding:

An acid to a base.

A base to an acid.

So if you recorded the pH of the mixture of the two solution as you mixed them you would get a titration curve.

pH

Volume of titrate added (mL)

If you added acid titrate to a base sample you would start at a high pH and it would gradually decrease as you added more acid, eventually leveling off when there’s lots of acid.

12-------------------------------------

2--------------------------------------

Pure Base

Pure Acid

7------------------------------------

Equivalence Point

Equivalence Point:

End Point:

Titrant:

Sample:

Is the point during a titration when the amount of acid (mols) is equal to the amount of base (mols).

Is the point during a titration a COLOUR change is observed due to the presence of an indicator added to the solution.

The solution in a titration that is added, AKA the solution in the big glass tube (burette).

Sodium Hydroxide titrated with Hydrochloric acid.

NaOH(aq) + HCl(aq) NaCl(aq) + HOH(l)1 1 1 1

Na+(aq) + OH-

(aq) H+(aq) + Cl-

(aq) Na+(aq) + Cl-

(aq) HOHl)

Total Ionic EquationOH-

(aq) + H+(aq)

HOHl)

Net Ionic Equation***This net ionic equation only has 1 H+

(aq) and 1 OH-(aq)***

pH

Volume of NaOH added (mL)

1. So before you begin there is just HCl, so you start the titration curve low.

2------------------------------------------

2. As you start adding NaOH the pH will start to rise as the HCl gets canceled out.

3. When all the HCl has been cancelled out, the next drop of NaOH that is added causes the pH to

jump. (mols are equal)

pH

Volume of NaOH added (mL)

2------------------------------------------

4. At the end of the tiration when there is only base left the pH will level off at the pure base level.

12-------------------------------------

7------------------------------------Equivalence Point

Sodium Carbonate titrated with Hydrochloric acid.

Na2CO3(aq) + HCl(aq) NaCl(aq) + H2CO3(aq)1 2 2 1

2 Na+(aq) + CO3

2-(aq) 2H+

(aq) + 2Cl-(aq) 2Na+

(aq) + 2Cl-(aq) H2CO3(aq)

Total Ionic EquationCO3

2-(aq) + 2H+

(aq) H2CO3(aq)

Net Ionic Equation***This net ionic equation has 2H+

(aq) and 1 CO32-

(aq)***

***Because there are MORE THAN 1 H+(aq) there will be

MORE THAN 1 “BUMP”.***

pH

Volume of HCl added (mL)

1. So before you begin there is just Na2CO3(aq), so you start the titration curve High.

2------------------------------------------

2. As you start adding HCl the pH will start to rise as the Na2CO3(aq) gets canceled out.3. When all the Na2CO3 has been cancelled out, the next drop of HCL that is added causes the

pH to jump. (mols are equal)

11-------------------------------------

pH

Volume of HCl added (mL)

4. BECAUSE there are 2 H+, there will be 3 levels instead of two.

2------------------------------------------

11------------------------------------- 4. Once all the intermediate NaCO3- get cancelled out and the pH drops again.

5. Once all the intermediate Na2CO3-AND NaCO3- get cancelled out and only HCl remains, the pH will level off at pure acid.

4------------------------------------Equivalence Point

End Points and Indicators***When choosing an indicator for a titration choose an indicator that will change colour as close to the equivalence pH as possible.***

Strong Acid With Strong Base 7 Equivalence Point

Strong Acid With Weak Base 5 Equivalence Point

Weak Acid With Strong Base 10 Equivalence Point

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