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Qualitative Analysis:. An analysis that determines what’s in a solution, the qualities of the solution. Quantitative Analysis:. An analysis that determines the amounts of a substance present in a solution. Qualitative Analysis. Examples:. 1. Solution Colour. Colorimetry :. - PowerPoint PPT Presentation

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Qualitative Analysis:Quantitative Analysis:An analysis that determines whats 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 Colour2. Flame Test

Data Booklet Pg. 6Quantitative AnalysisExamples:1. Gravimetric AnalysisThe process of using stoichiometry to calculate masses of unknown substances.

Gravimetric (solids)StoichiometryGas Stoichiometry

Solution StoichiometryQuantitative AnalysisExamples:1. Gravimetric Analysis2. Titration Analysis

An 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 + AgClPredict the products and balance.1111

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 + AgCl1111Now use the solubility chart to determine the statessoluble (aq) or slightly soluble (s) (aq)(aq)(aq)(s)PrecipitateAgNO3 + NaCl NaNO3 + AgCl1111(aq)(aq)(aq)(s)PrecipitateWhen 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 EquationCu +AgNO3Cu(NO3)2 +AgStates?(s)(aq)1122(s)(aq)Example

2. Write down what you know.Cu +AgNO3Cu(NO3)2 +Ag(s)(aq)1122(s)(aq)m = 10.0 gm = 20.0 gExample

3. Convert all quantities to mols.Cu +AgNO3Cu(NO3)2 +Ag(s)(aq)1122(s)(aq)m = 10.0 gm = 20.0 g

M = 63.55 g/moln = 0.157molM = 169.88 g/moln = 0.118 molExample

4. Convert one reagent into the other using N/G and see which is larger/bigger.Cu +AgNO3Cu(NO3)2 +Ag(s)(aq)1122(s)(aq)m = 10.0 gm = 20.0 gM = 63.55 g/moln = 0.157molM = 169.88 g/moln = 0.118 mol( )NG21(0.157 mols)(2/1) =0.317 molBecause more mols of AgNO3 can be produced than are actually present, AgNO3 is the limiting reagent.L.RE.R

ExampleCu +AgNO3Cu(NO3)2 +Ag(s)(aq)1122(s)(aq)m = 10.0 gm = 20.0 gM = 63.55 g/moln = 0.157molM = 169.88 g/moln = 0.118 molL.RE.ROnce 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 molm = ?m = 12.7 g

ExampleCu +AgNO3Cu(NO3)2 +Ag(s)(aq)1122(s)(aq)m = 10.0 gm = 20.0 gM = 63.55 g/moln = 0.157molM = 169.88 g/molL.RE.RYou 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 molm = 12.7 g( )12 n = 0.0589 molnCu(s) Left Over= Mols of E.R mols from L.R= 0.157mol 0.0589mol= 0.0981n = 0.0981

ExampleCu +AgNO3Cu(NO3)2 +Ag(s)(aq)1122(s)(aq)m = 10.0 gm = 20.0 gM = 63.55 g/moln = 0.157molM = 169.88 g/molL.RE.RYou can calculate the MASS of Cu left over by converting the mols into grams by multiplying by the molar mass.n = 0.118 molm = 12.7 gn = 0.0981Mass Cu(s) = nM = (0.0981 mol)(63.55 g/mol) = 6.23 gExample 2

FeCl3(aq) + NaOH(aq) Fe(OH)3(aq) + NaCl(aq)1331m = 26.8 gm = 21.5 gn = m Mn = 0.165 moln = 0.538 mol( )31 n = 0.496 molL.RE.R0.538mol 0.496 mol = mols of NaOH left over.0.042 mol NaOH left over.Example 2

FeCl3(aq) + NaOH(aq) Fe(OH)3(aq) + NaCl(aq)1331m = 26.8 gm = 21.5 gn = 0.165 moln = 0.538 molL.RE.R0.042 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

FeCl3(aq) + NaOH(aq) Fe(OH)3(aq) + NaCl(aq)1331m = 26.8 gm = 21.5 gn = 0.165 moln = 0.538 molL.RE.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 molm = nMm = 17.7 gExample 2

FeCl3(aq) + NaOH(aq) Fe(OH)3(aq) + NaCl(aq)1331m = 26.8 gm = 21.5 gn = 0.165 moln = 0.538 molL.RE.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 = nMm = 17.7 gn = 0.495 molm = 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)1221

Na2CO3(aq)+HCl(aq) H2CO3(aq)+NaCl(aq)1221Step 2: Write down what you know.n = m MV = 0.100 LC = ?m = 1.59 gn = 1.59 g 105.99 g/mol

n = 0.01500142 moln = 0.0150 mol

Na2CO3(aq)+HCl(aq) H2CO3(aq)+NaCl(aq)1221Step 2: Write down what you know.V = 0.100 LC = ?n = 0.0150 molC = n V= 0.0150 mol 0.100LC = 0.150 mol/LC = 0.150 mol/L

Na2CO3(aq)+HCl(aq) H2CO3(aq)+NaCl(aq)1221Step 2: Write down what you know.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 didnt 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.

Na2CO3(aq)+HCl(aq) H2CO3(aq)+NaCl(aq)1221Step 2: Write down what you know.C = ?C = 0.150 mol/LV = 0.010 LV = 0.0128 LFrom now on its just solution stoichiometry.n = CVn = 0.0015 mol( )21 n = 0.003 mol

Na2CO3(aq)+HCl(aq) H2CO3(aq)+NaCl(aq)1221Step 2: Write down what you know.C = ?C = 0.150 mol/LV = 0.010 LV = 0.0128 Ln = 0.003 molC = 0.003 mol 0.0128 LC = n VC = 0.234 mol/LC = 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.pHVolume 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 theres lots of acid.12-------------------------------------2--------------------------------------

Pure BasePure Acid7------------------------------------Equivalence PointEquivalence 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)1111Na+(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)***

pHVolume 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