ch3 analytical and quantitative chemistry

7/30/2019 CH3 Analytical and Quantitative Chemistry

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Chemical Testing Ionic SubstancesIonic substances contain positive and negative ions, we use different tests to identify the

positive and the negative ions:

red orange yellow apple green blue/green lilac

Identifying some positive (metal) ions

e.g. Na+, Mg2+ etc.

The first test to do is a flame tests some

metal ions give characteristic flame colours:

If our unidentified substance produces no flame colour, the next test is to make

a solution containing the ions of our unknown, and to add sodium hydroxide

solution. Many substances form precipitates of metal hydroxides which havecharacteristic properties or colours:

Ion formula effect of adding sodium hydroxide, NaOH

aluminium Al3+ white precipitate, which re-dissolves in excess NaOH

calcium Ca2+ white precipitate , does not dissolve in excess NaOH

magnesium Mg2+ white precipitate, does not dissolve in excess NaOH

copper Cu2+ pale blue precipitate

iron(II) Fe2+ dark green precipitate

iron(III) Fe3+ orange-brown precipitate

Note: calcium and magnesium ions both give identical precipitates, but they can be told apartbecause calcium ions produce a flame colour but magnesium ions do not.

Cu2+

Fe2+

Fe3+


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Identifying some negative ions- e.g. halide ions Cl-, Br-, I-, sulphate ions SO4

2-, carbonate ions CO32-

We use a sequence of chemical tests to determine which negative ion is present:

1: Carbonates fizz when an acid, e.g. hydrochloric acid, is added,because carbon dioxide gas is given off. This can be identified bybubbling it through limewater, which turns cloudy.

2: Solutions containing halide ions produce precipitates of silver

halides when treated with nitric acid and silver nitrate- silverchloride is WHITE- silverbromide is CREAM- silveriodide is YELLOW.

The nitric acid is added first to remove interfering ions such as

hydroxide or carbonate.

3: Sulphate ions can be identified by adding barium chloride whichhas been acidified with hydrochloric acid. A white precipitate of

barium sulphate is produced.

The acid is added first to remove interfering ions such as hydroxide

or carbonate.

Cl- Br- I-


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Practice Questions:

1) Substance A is ionic. It produces no flame colour, but does produce a dark green precipitate

when sodium hydroxide is added. If nitric acid is added to substance A, followed by silver

nitrate solution, a white precipitate is seen. What is A ?

2) Substance B is ionic. A pale apple-green flame is produced during a flame test. When

hydrochloric acid is added to B fizzing is seen and the gas given off turns limewater cloudy.

What is B ?

3) Substance C is sodium sulphate. How would you prove this using chemical tests ?

4) Three substances have lost their labels. Each is a white ionic compound. One label reads

Aluminium bromide one label reads magnesium bromide and the third reads calcium

bromide. Explain how you could use simple laboratory tests to decide which label to stick on

which substance.


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The units of concentration are moles per cubic decimetre.

We write this as mol/dm3 or mol dm-3 Note: 1dm3 = 1000cm 1 dm3 = 1 litre

An acid solution with a concentration of1 mol/dm3 has one mole of acid particles dissolved in1 dm3 of the solution.

A solution of0.1 mol/dm3 is only a tenth of the concentration, i.e. it is ten times more dilute.

We can calculate the concentration of a solution if we know how many moles of solute are

there, and what volume of solution they are dissolved in:

concentration (mol /dm3) = moles volume (in dm3)

e.g. 7.3g of HCl are dissolved in 0.1 dm3(100cm3) of water. What is the concentration of theHCl solution ?

moles = mass RFM = 7.3 36.5 = 0.2 molesconcentration = moles volume = 0.2 0.1 = 2 mol/dm3

Dilute:a dilute acid (or alkali) has a small number

of acid molecules per cm3 of aqueous

solution.

Working with solutions

Concentrated:a concentrated acid (or alkali) has a largenumber of acid molecules per cm3 of

aqueous solution.


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We can also work out how many moles are in a solution if we know its concentration and its

volume:

moles = concentration (mol/dm3) x volume (dm3)

e.g. How many moles of sodium hydroxide (NaOH) would I need to dissolve to make up

0.5 dm3(500cm3) of solution with 0.1 mol/dm3concentration ?

moles of NaOH = concentration x volume = 0.1 x 0.5 = 0.05 mol

What would be the mass of sodium hydroxide I should weigh out to make up this solution ?The RFM of NaOH = 23 (Na) + 16 (O) + 1(H) = 40mass of NaOH = moles of NaOH x RFM of NaOH = 0.05 x 40 = 2.0g

Practice Questions:

5) How many moles of sodium chloride would I need in order to make 250cm3 of solution with

concentration 2 mol/dm3 ?

6) What mass of sodium chloride would be needed to make this solution ?

7) If I only had 5.85 g of sodium chloride, what concentration solution would I get if I dissolved it

to make 100cm3 of solution ?


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TitrationTitration is a technique used to measure how much of an acid is

needed to exactly neutralise an alkali.

If we know the concentration of either the acid or the alkali, we

can use titration to find the concentration of the other.

The experiment:An indicator is mixed with a known volume of alkali in the flask.

The level of the acid in the burette is noted, then acid is added

from the burette into the flask, carefully, dropwise, until the

indicator just changes colour. This is called the endpoint.

The acid level is noted again, and the volume of acid that has

been added is worked out.

Repeats are done to get consistent results which can be

averaged.

some common indicators showing the colour change at the endpointN.B. Universalindicator is notsuitable for titrationas it has manycolour changes andhence no clear

endpoint


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Finding the concentration of an acid:e.g. Sulphuric acid of unknown concentration was neutralised in a titration experiment using

sodium hydroxide solution of 0.10 mol/dm3 concentration: H2SO4 + 2 NaOH Na2SO4 + H2O

The conical flask contained 25.0cm3 of NaOH solution and the indicator changed colour after

16.5cm3 of sulphuric acid had been added. Work out the concentration of the acid.

Step 1:Work out the number of moles of alkali (NaOH) in the flask

moles of NaOH = concentration of NaOH x volume of NaOH in dm3 (25cm3= 0.0250dm3)= 0.10 x 0.0250= 0.00250 moles of NaOH

Step 2:Use the mole ratio from the chemical equation to work out how many moles of sulphuric acid it

took to neutralise the alkali.

mole ratio = 1 H2SO4 : 2 NaOH (from balanced equation)= 0.00125 : 0.00250 molesso moles of sulphuric acid needed = 0.00125

Step 3:Work out the concentration of the acid (H2SO4)

conc. of H2SO4 = moles of H2SO4 volume of H2SO4 in dm3 (16.5cm3= 0.0165dm3)

= 0.00125 0.0165= 0.075 mol/dm3


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Finding the concentration of an alkali:THE STEPS ARE THE SAME WITH ACID AND ALKALI SWAPPED OVER

1) work out the moles of acid used

2) use the mole ratio to work out moles of alkali used

3) convert moles of alkali to concentration of alkali

You might find a table-format more helpful when doing titration calculations. The example

answer below is exactly the same method, just laid out differently:

e.g. In a titration, 25.0cm3 of NaOH solution of unknown concentration was neutralised. The

indicator changed colour after 20.0cm3 of HCl of 0.1 mol/dm3 concentration had been

added. Work out the concentration of the alkal. HCl + NaOH NaCl + H2O

equation HCl + NaOH NaCl + H2O

concmol/dm3

0.1 = 0.002 0.025

= 0.08

volume dm3 20cm3 =0.020 dm3

25cm3 =0.025 dm3

moles = 0.1 x0.020= 0.002

0.002

mole ratio 1 1 1 1

Answer: The concentration of the alkali was 0.08 mol/dm3


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Answers to Practice questions:

1) Dark green precipitate with sodium hydroxide identifies iron(II) ions

White precipitate with acidified silver nitrate identifies chloride ions

A is iron(II) chloride

2) Apple-green flame colour identifies barium ions

Fizzing when acid added identifies carbonate ions

B is barium carbonate

3) The presence of sodium ions can be shown using a flame test yellow flame colour

sulphate ions give a white precipitate if the sample is acidified with hydrochloric acid

then barium chloride solution added

4) A flame test will identify the calcium bromide orange/brick red flame colour

whereas the other two will produce no flame colour. Adding sodium hydroxide to

solutions of each of the two remaining compounds will produce white precipitates in

both cases, but if more sodium hydroxide is added (an excess) the precipitate

produced by the aluminium ions will re-dissolve leaving a colourless solutionwhereas the magnesium bromide precipitate will not re-dissolve in excess sodium

hydroxide.


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5) moles of NaCl = concentration of NaCl x volume of solution in dm3 (250cm3 = 0.25 dm3)

= 2 x 0.25 = 0.50 moles

6) mass of NaCl = moles of NaCl x RFM of NaCl RFM = 23(Na) + 35.5(Cl) = 58.5

= 0.50 x 58.5 = 29.25g

7) moles of NaCl = mass of NaCl RFM of NaCl

= 58.5 / 58.5 = 1.00 moleconcentration of solution = moles of NaCl / volume of solution in dm3 (100cm3 = 0.1dm3)

= 1.00 x 0.1 = 0.1 mol/dm3

ch3 analytical and quantitative chemistry

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