Acid-base titrations

-OH

-OH

-OH

-OH

Titration introduction

• Purpose : neutralization reaction - quantitative measure (how much acid, how much base?)

• Setup : burette clamps, retort stands.

• Equipment : rinsing procedure

• Pipettes : review how to measure 10 cm.

• Burettes : expensive, differences, clogged tips

• Titration : reading volumes, no air in tip, repeat

• Chemicals: Never return solutions to container

3

Sources of errorsOvershoots/misses end-point Water left in burette / pipette Air lock below tap in burette / air in pipette Burette not vertical Alkali not at stated concentration leaking tap Not reading meniscus at eye-level Funnel left in top of burette Not reading level against a white background Not reading meniscus correctly Washing pipette between titres Washing the flask with the solution that will go in it Not swirling flask / mixture

Minimise sources of errors•use white tile or paper as background to burette readings•have eyes level with meniscus (to avoid parallax error)•measure level at bottom of the meniscus•ensure that the burette is upright•ensure that there are no air bubbles•ensure that the jet is full•remove the funnel from the top of the burette•use a white tile beneath flask•swirl the (conical) flask•run solution from burette slowly (into conical flask)•ensure no solution from the burette is left on the sides/walls of the (conical) flask•add solution from burette drop-wise close to the end-point

28

27

27.95 cm3

Observed

True value 27.95 cm3

+0.0528.01

True value 27.95 cm3

- 0.0527.90

987654321

Percentage error0.05 X 2 X 100 27.95=0.357%

Concordant resultsThe titres values within 0.2 (cm3) orwithin 0.1 (cm3) consistent

Titration number 1 2 3 4Burette reading

(final) / cm3

28.55 28.00 40.35 28.05

Burette reading (initial) / cm3

0.00 0.05 12.30 0.05

Volume of NaOH used / cm3

28.55 27.95 28.05 28.00

27.95+28.05+28.00 = 28 3

mean/average titre value =

23. Methyl orange is red in acidic solutions and yellow in alkaline solutions. What is the colour of the indicator at theend point of a titration of aqueous sodium hydroxide solution with hydrochloric acid?A red B pink C orange D yellow

Copper(ii) thiosulphate titrationsCopper(ii) thiosulphate titrations

General theoryGeneral theoryCopper(II) compounds can be analysed by a redox titration.

The general procedure is that excess potassium iodide solution is added to a neutral solution of copper(II). This liberates iodine according to the equation below and the amount of iodine is found by titration with sodium thiosulphate solution. Just before the end-point, several drops of starch solution are added and one continues the titration until the blue colour just disappears and an off-white precipitate remains.

2Cu2+(aq) + 4I¯(aq) 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) S4O62-(aq) + 2I¯(aq)

therefore moles of S2O32- = moles of Cu2+(aq)

Practical details

Pipette a known volume of a solution of Cu2+ ions into a conical flask.(alternatively dissolve a known mass of solid in water)

Neutralise the solution by adding sodium carbonate solution dropwise until a feint precipitate starts to form.

A

A

Practical details

Add excess potassium iodide solution to liberate iodine. The copper(II) is reduced to copper(I) and half the iodide ions are oxidised to iodine.

2Cu2+(aq) + 4I¯(aq) 2CuI(s) + I2(aq)off white solid

A B

B

+2 -1 +1 -1 0

Practical details

Titrate with a standard solution of sodium thiosulphate until the solution lightens. DO NOT ADD TOO MUCH.The iodine is reduced back to iodide ions.

2S2O32-(aq) + I2(aq) S4O6

2-(aq) + 2I¯(aq)

A B C

C

0 -1

Practical details

Starch solution is added near the end point.

Starch gives a dark blue colouration in the presence of iodine.

A B C D

D

Practical details

Continue with the titration, adding the sodium thiosulphate dropwise until the blue colour disappears at the end point. This indicates that all the iodine has reacted.

Record the volume added and repeat to obtain concordant results.

A B C D E

E

State the colours of the titration solution just before the starch solution is added, afterthe starch solution is added and the colour change at the end-point of the reaction.??

Colour just before adding thestarch:(very) pale yellow/straw coloured

Colour after adding the starch:Blue-black

Colour at the end point:colourless

TYPICAL CALCULATIONSTYPICAL CALCULATIONS

Percentage copper in a compound

1 titrate a known mass of copper compound or a known volume of a solution

2 calculate the moles of S2O32- needed

3 according to the equation… moles of Cu2+ = moles of S2O32-

4 calculate the number of moles of Cu2+

5 calculate the mass of copper by multiplying the moles of copper by the molar mass of copper.

6 divide the mass of copper by the mass of the weighed solid to find the fraction and hence calculate the percentage of copper in the sample.

Typical calculationTypical calculation

Number of water molecules of crystallisation1 titrate a known mass of copper compound or a known volume of a solution2 calculate the moles of S2O3

2- needed3 according to the equation… moles of Cu2+ = moles of S2O3

2-

4 calculate the number of moles of Cu2+

5 calculate the number of moles of CuSO4

moles of CuSO4 = moles of Cu2+ (there is one Cu2+ in every CuSO4)6 calculate the mass of copper sulphate by multiplying the moles of copper sulphate by the molar mass of copper sulphate (CuSO4)7 calculate mass of water(= mass of CuSO4.xH2O - mass of CuSO4)8 divide the mass of water by 18 to find the number of moles of water9Compare the ration of moles of… water : moles of CuSO4 to find x

3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a 0.100 mol dm sodium thiosulphate solutionCalculate the percentage of copper in the alloy.

Calculation – Example 1Calculation – Example 1

3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a 0.100 mol dm sodium thiosulphate solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq) + 4I¯(aq) 2CuI(s)+ I2(aq)2S2O3

2-(aq) + I2(aq) S4O62-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

Calculation – Example 1Calculation – Example 1

3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a 0.100 mol dm sodium thiosulphate solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq)+4I¯(aq) 2CuI(s)+I2(aq)

2S2O32-(aq) + I2(aq) S4O6

2-(aq) + 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.100 x 25.00 / 1000 = 2.50 x 10-3

Calculation – Example 1Calculation – Example 1

3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a 0.100 mol dm sodium thiosulphate solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq)+4I¯(aq) 2CuI(s)+ I2(aq)2S2O3

2-(aq) + I2(aq) S4O62-(aq)+ 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.100 x 25.00 / 1000 = 2.50 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.50 x 10-3

Calculation – Example 1Calculation – Example 1

3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a 0.100 mol dm sodium thiosulphate solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq) + 4I¯(aq) 2CuI(s)+ I2(aq) 2S2O3

2-(aq) + I2(aq) S4O62-(aq)+ 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.100 x 25.00 / 1000 = 2.50 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.50 x 10-3

moles of Cu2+ in 250cm3 = 2.50 x 10-3 x 10= 0.025

Calculation – Example 1Calculation – Example 1

-3

3.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a 0.100 mol dm sodium thiosulphate solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq)+4I¯(aq) 2CuI(s) + I2(aq)2S2O3

2-(aq) + I2(aq) S4O62-(aq)+ 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.100 x 25.00 / 1000 = 2.50 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.50 x 10-3

moles of Cu2+ in 250cm3 = 2.50 x 10-3 x 10= 0.025

mass of Cu2+ in 250cm3 = 0.025 x 63.5 = 1.588g

Calculation – Example 1Calculation – Example 1

-3

Calculation – Example 1Calculation – Example 13.00g of a copper alloy was dissolved in acid and the solution made up to 250cm3. 25cm3 was then pipetted into a conical flask and an excess of KI added. It was found 25cm3 of a 0.100 mol dm sodium thiosulphate solutionCalculate the percentage of copper in the alloy.

From these two equations 2Cu2+(aq)+4I¯(aq) 2CuI(s)+ I2(aq) 2S2O3

2-(aq) + I2(aq) S4O62-(aq)+ 2I¯(aq)

you get moles of S2O32- = moles of Cu2+(aq)

moles of S2O32- = 0.100 x 25.00 / 1000 = 2.50 x 10-3

moles of Cu2+ in 25cm3 = moles of S2O32- = 2.50 x 10-3

moles of Cu2+ in 250cm3 = 2.50 x 10-3 x 10= 0.025

mass of Cu2+ in 250cm3 = 0.025 x 63.5 = 1.588g

% of Cu in the alloy = 1.588 / 3.00 x 100 = 52.91%

-3

Calculate the volume of dilute hydrochloric acid, concentration 0.200 mol dm , needed to neutralize 20 cm of aqueous calcium hydroxide, concentration 0.100 mol dm .2HCl(aq) + Ca(OH)2(aq) CaCl2(aq) + 2H2O(l)A 10 cm B 20 cm C 40 cm D 80 cm

During a titration, when the solution in a pipette is transferred to a conical flask, a small amount of liquid remains in the tip of the pipette. This situation should be dealt with byA leaving the liquid in the pipette which is calibrated to allow for it.B slightly over-filling the pipette to compensate for the additional volume.C carefully blowing the liquid out of the pipette to ensure that it is empty.D repeating the titration.

-3

3333

3

The tolerance of a 25 cm pipette is ±0.06 cm . The percentage error in the measurement of 25 cm using this pipette isA ±0.06%B ±0.12%C ±0.24%D ±0.48%

A series of titrations is carried out using the same conical flask. Before carrying out each titration, the conical flask must beA rinsed with ethanol.B rinsed with distilled or deionised water.C rinsed with the solution that it will contain.D dried to remove all traces of liquid.

33

3

Weak acids such as ethanoic acid cannot be titrated with weak bases such as ammonia using an indicator since there is never any distinct colour change.An alternative technique is to use thermometric titration as follows.1. 30.0 cm of dilute ethanoic acid is placed in a polystyrene cup and its temperature measured.2. 5.00 cm of ammonia solution of concentration 1.05 mol dm–3 is then added to the acid, the mixture stirred and the temperature measured again.3. Further 5.00 cm portions of ammonia are added, followed by measurement of the temperature, until a total of 35.0 cm has been added.The results of this experiment are tabulated below.

3

3

3

3

(a) (i) Plot these data on the axes below. Draw two straight lines through the points on your graph. Extrapolate the lines until they intersect, to enable you to determine the end-point volume.

Volume of ammonia / cm

o

3

(ii) State the volume of the ammonia solution at the end-point.(iii) Explain why the temperature rises until the end-point is reached.(iv) Explain why the temperature falls when more ammonia solution is added after the end-point.(b) In a similar experiment, 25.0 cm of ethanoic acid of concentration 2.00 mol dm was reacted with 25.0 cm of 2.00 mol dm aqueous ammonia. The initial temperature was 20.6 °C and the temperature at the end-point was 29.8 °C.(i) Use the expression below to calculate the heat energy evolved in this reaction. (Assume that the density of the reaction mixture is 1.00 g cm and that the specific heat capacity of the mixture is 4.18 J g °C .) energy transferred = mass × specific heat capacity × temperature change in joules

3

-3

-3

-3 3

-1 -1

(ii) Calculate the number of moles of ethanoic acid used in this reaction.(iii) The reaction that occurs isCH3COOH(aq) + NH3(aq) → CH3COONH4(aq)Use your values from (b)(i) and (ii) to calculate the enthalpy change per mole for this reaction. Include a sign and units in your answer. Give your answer to three significant figures.

The enthalpy change of neutralization of an acid by an alkali is measured by adding 10.0 cm3 of hydrochloric acid to 10.0 cm3 of sodium hydroxide. 10.0 cm3 pipettes with an accuracy of ±0.04 cm3 are used to measure out both solutions.The overall percentage error in measuring the total volume of the reaction mixture isA ±0.04% B ±0.08% C ±0.4% D ±4.0%