Volumetric Tritration Calculation

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12. Molarity, moles & mass and Volumetric titration calculations e.g. acid-alkali titrations

Titrations can be used to find the concentration of an acid or alkali from the relative volumes used and the concentration of one of the two

reactants. Themethod and apparatusused are briefly described at the end of this page.

You should be able to carry out calculations involving neutralisation reactions in aqueous solution given the balanced equation or from

your own practical results.

The examples insection 7. molesand mass. andsection 11. concentrationwill help you follow the

calculations below.

o Note again: 1dm3

= 1 litre = 1000ml = 1000 cm3, so dividing cm

3/1000 gives dm

3.

o and other useful formulae or relationships are:

moles = molarity (mol/dm3) x volume (dm3 = cm3/1000),

molarity (mol/dm3) = mol / volume (dm

3= cm

3/1000),

1 mole = formula mass in grams.

o In most volumetric calculations of this type, you first calculate the known moles of one reactant from a

volume and molarity. Then, from the equation, you relate this to the number of moles of the other reactant, and then with

the volume of the unknown concentration, you work out its molarity.

Example 12.1: Given the equation NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l)

o 25.0 cm3

of a sodium hydroxide solution was pipetted into a conical flask and titrated with 0.200 mol

dm-3 (0.2M) hydrochloric acid.

o Using a suitable indicator it was found that 15.0 cm3

of the acid was required to neutralise the alkali.

Calculate the molarity of the sodium hydroxide and its concentration in g/dm3.

moles = molarity x volume (in dm3

= cm3/100)

moles HCl = 0.200 x (15.0/1000) = 0.003 mol

moles HCl = moles NaOH (1 : 1 in equation)
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There are more questions involving molarity insection 11. introducing molarityandsection 14.3 on dilution

How to carry out a titration?

The right diagrams show the typical apparatus (1)-(6) used in manipulating liquids and on the left a brief three stage description

of titrating an acid with an alkali:

1.

An accurate volume of acid is pipetted into the conical flasks using a suction bulb and pipette for health and safetyreasons. Universal indicatoris then added, which turns red in the acid.

2. The alkali, of known accurate concentration, is put in the burette and you can conveniently level off the

reading to zero (the meniscus on the liquid surface should rest on the zero -- graduation mark).

Note in stage 2. other possibilities are:

A small amount of accurately weighed solid acid is dissolved in water and titrated with alkali.
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A small amount of accurately weighed solid alkali is dissolved in water and titrated with acid.

After this, the method is essentially the same as described below.

3.

The alkali is then carefully added by running it out of the burette in small quantities, controlling the flow with the tap, until the

indicator seems to be going yellow-pale green.

The conical flask should be carefully swirled after each addition of alkali to ensure all the alkali reacts.

4. Near the end of the titration, the alkali should added drop-wise until the universal indicator goes green.

This is called the end-point of the titration and the green means that all the acid has been neutralised.

The volume of alkali needed to titrate-neutralise the acid is read off from burette scale, again reading the volume value on

the underside of the meniscus.

The calculation can then be done to work out the concentration of the alkali.

5. Universal indicator, and most other acid-base indicators, work for strong acid and alkali titrations, but universal indicator isa somewhat crude indicator for other acid-alkali titrations because it gives such a range of colours for different pH's. Examples of

more accurate and 'specialised' indicators are:

titrating a strong alkali with a strong acid (or vice versa):

e.g. for sodium hydroxide (NaOH) - hydrochloric/sulphuric acid (HCl/H2SO4) titrations, use ...

phenolphthalein indicator (pink in alkali, colourless in acid-neutral solutions), the end-point is the pink

colourless change.

Litmus works too, the end point is the red purple/blue colour change.

titrating a weak alkali with a strong acid:

e.g. for titrating ammonia (NH3) with hydrochloric/sulfuric acid (HCl/H2SO4), use ...

methyl orange indicator (red in acid, yellowish-orange in neutral-acid), the end-point is an 'orange' colour, not

easy to see accurately.

screened methyl orange indicator is a slightly different dye-indicator mixture that is reckoned to be easier to see

than methyl orange, the end-point is a sort of 'greyish orange', but still not easy to do accurately.


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titrating a weak acid with a strong alkali :

e.g. for titrating ethanoic acid (CH3COOH) with sodium hydroxide (NaOH), use ...

phenolphthalein indicator (pink in alkali, colourless in acid-neutral solutions, pink in alkali), the end-point is the first

permanent pink.

methyl red indicator (red in acid, yellow in neutral-alkaline), the end-point is 'orange'.

titrating a weak acid with a weak alkali (or vice versa):

These are NOT practical titrations because the pH changes at the end-point are not great enough to give a

sharp colour change with any indicator.

TheAcids, Bases, pH page section (2) lists common indicators.

The theory, and examples of strong/weak acids/alkalis (soluble bases) are described on theExtra Aqueous Chemistry

page section 3,

and theAcids, Bases, pH page section (7) explains the changes in pH in the titration.

Advanced level theory of indicators and titrationsandadvanced acid-alkali titration questions

6.1 Salt Hydrolysis, acidity and alkalinity of salt solutions

Despite being taught at lower academic levels that salts e.g. sodium chloride, dissolve in water to form neutral solutions of pH 7.

In reality, and looking at a wider variety of 'salts', the picture is much more complicated and a 'salt' solution may be acid, neutral or

alkaline depending on the nature of the interaction of the salt ions with water.

The reasons are quite clear when you consider the possible Bronsted-Lowry interactions that can take place between the ions of the salt

and water.

6.1.1 Examples of acidic salt solutions: pH


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or more simply: [Al(H2O)6]3+

(aq) [Al(H2O)5(OH)]2+

(aq) + H+

(aq)

The hydrated metal ion acts as an acid (proton donor) and water acts as the base (proton acceptor) and so aqueous

hydrogen/oxonium ions are formed.

The greater the charge on the central metal ion, the stronger the hexa-aqua ion acid. e.g.

[Al(H2O)6]3+

(aq)> [Mg(H2O)6]2+

(aq)> [Na(H2O)6]+

(aq) (the cations of Gps 1-3 on Period 3)

or[Fe(H2O)6]3+

(aq)> [Fe(H2O)6]2+

(aq)(in the 3d-block transition metal example)

From left to right, the trend is due to a decreasing charge density effect of the central metal ion on the O-H bond

of a co-ordinated water molecule. The charge density decreases as the positive charge of the central metal ion

decreases and its ionic radius increases.

The sodium ion shows virtually no acidic behaviour.

Further discussion of this situation will be on the Transition Metals Appendix page section 1 (currently under

production).

However, the anion of the salt must not be neglected for a full explanation. The anions derived from the very strong

hydrochloric/sulphuric/nitric acids are all very weak bases and so have little tendency to interact with water in an acid-

base reaction. Its a general, and logical rule, that the conjugate base of a very strong acid is very weak.

o 6.1.1b: Salts of weak bases and strong acids give acidic solutions.

e.g. ammonium chloride. The chloride ion is such a weak base that there is no acid-base reaction with water, but the

ammonium ion is an effective proton donor. As a general rule, the conjugate acid of a weak base is quite strong. The

result here is that ammonium salt solutions have a pH of 3-4.

NH4+

(aq) + H2O(l) NH3(aq) + H3O+

(aq)

In zinc-carbon batteries an acidic ammonium chloride paste dissolves the zinc in the cell reaction, though an oxidising

agent must be added (MnO2) to oxidise the hydrogen formed into water, or batteries might regularly explode!

If you place a piece of magnesium ribbon or a zinc granule in ammonium chloride or ammonium

sulphate solution you will see fizzing as hydrogen gas is formed.
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2H3O+

(aq) + M(s)==> M2+

(aq) + H2O(l) + H2(g)

M = zinc or magnesium

6.1.2 Examples of nearly neutral salt solutions: pH approx. 7

o 6.1.2a: Salts of strong acids and strong bases e.g. sodium chloride

Here the hexa-aqua sodium ion shows no acidic behaviour and the chloride ion no base behaviour, so little or no

interaction with water to produce either H+

(aq) or OH-(aq) to change the pH.

o 6.1.2b: Salts of weak acids and weak bases: e.g. ammonium ethanoate

Here the ammonium ion can act as an acid to form H+

(aq) with water, but the ethanoate ion acts as a base to give OH-

(aq) with water, so they effectively neutralise each other.

6.1.3 Examples of alkaline salt solutions: pH>7

o 6.1.3a: Salts of a weak acid and a strong base e.g. sodium ethanoate

The hydrated sodium ion shows no acidic character but the ethanoate ion is a strong conjugate base of a the weak

ethanoic acid (pKa = 4.76, Ka = 1.74 x 10-5

mol dm-3

), so an acid-base hydrolysis reaction occurs to generate hydroxide

ions to raise the pH to about pH 9.

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH

-(aq)

o 6.1.3b: Potassium cyanide: is the salt of the very strong base potassium hydroxide and the very weak hydrocyanic acid (pKa =

9.31, Ka = 4.9 x 10-10

mol dm-3

). The hydrated potassium ion shows no acidic behaviour, but the cyanide ion is a strong conjugate

base of the very weak hydrocyanic acid (HCN) which interacts with water to generate hydroxide ions. Hydrocyanic acid (pKa = 9.4)

is weaker than ethanoic acid (pKa = 4.76) , so the equilibrium is more on the right, more OH-, and so the pH is more alkaline, i.e.

over 9.

CN-(aq) + H2O(l) HCN(aq) + OH

-(aq)

o 6.1.3c: Sodium carbonate is the 'salt' of the strong base sodium hydroxide and the very weak 'carbonic acid'

Again the hydrated sodium ion shows no acidic character but the carbonate ion is a strong conjugate base of a the weak

'carbonic' acid, so an acid-base hydrolysis reaction occurs to generate hydroxide ions to raise the pH.


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CO32-

(aq) + H2O(l) HCO3-(aq) + OH

-(aq)

Volumetric Tritration Calculation

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