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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. The method and apparatus used
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 in
section 7. moles and mass. and
section 11. concentration will
help you follow the calculations below.
-
Note again:
1dm3 = 1 litre = 1000ml = 1000 cm3, so dividing
cm3/1000 gives dm3.
-
and other useful formulae or relationships are:
- moles = molarity
(mol/dm3) x volume (dm3 = cm3/1000),
- molarity (mol/dm3)
= mol / volume (dm3 = cm3/1000),
- 1 mole = formula mass
in grams.
- 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)
- 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.
- 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)
- so there is 0.003 mol NaOH in 25.0 cm3
- scaling up to 1000 cm3
(1 dm3),
there are ...
- 0.003 x (1000/25.0) = 0.12 mol NaOH in 1 dm3
- molarity of NaOH is 0.120 mol
dm-3 (or 0.12M)
- since mass = moles x formula mass, and Mr(NaOH)
= 23 + 16 + 1 = 40
- concentration in g/dm3
= molarity x formula mass
- concentration in g/dm3
is 0.12 x 40 = 4.80 g/dm3
- Example 12.2: Given the equation
2KOH(aq)
+ H2SO4(aq) ==> K2SO4 + 2H2O(l)
- 20.0 cm3 of a sulphuric acid solution was
titrated with 0.0500 mol dm-3 (0.05M) potassium hydroxide.
- If the acid required 36.0 cm3
of the alkali KOH for neutralisation what was the concentration of the acid?
- moles = molarity x volume
(in dm3 = cm3/100)
- mol KOH = 0.0500 x (36.0/1000) = 0.0018 mol
- mol H2SO4
= mol KOH / 2 (because
of 2 : 1 ratio in equation above)
- mol H2SO4
= 0.0018/2 = 0.0009 (in
20.0 cm3)
- scaling up to 1000 cm3
of solution = 0.0009
x (1000/20.0) = 0.0450 mol
- mol H2SO4 in 1 dm3 =
0.0450
- so molarity of H2SO4
= 0.0450 mol dm-3 (0.045M)
- since mass = moles x formula mass, and Mr(H2SO4)
= 2 + 32 + (4x16) = 98
- concentration in g/dm3
is 0.045 x 98 = 4.41 g/dm3
- Example 12.3:
There are more questions
involving molarity in section 11.
introducing molarity and
section 14.3
on dilution

- 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:
- (i) An accurate volume of acid is
pipetted into the conical flasks using a suction bulb and pipette for health and safety
reasons. Universal indicator is then added, which turns red in the
acid.
-
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 other
possibilities are:
- (ii) An accurate volume of alkali is
measured into a flask and titrated with an acid solution of known
concentration.
- (iii) A small amount of accurately
weighed solid acid is dissolved in water and titrated with alkali.
- (iv) A small amount of accurately
weighed solid alkali is dissolved in water and titrated with acid.
- This can procedure can be
used to compare the effectiveness of ant-acid indigestion
tablets, which are designed to neutralise excess acid in the
stomach.
- A known and equal mass of each
brand of indigestion tablet is crushed and mixed with some water
eg 20 cm3 (fair test points).
- Make sure the mixture is gently
swirled to completely dissolve the crushed tablet powder.
- The burette is filled with a
standard solution of hydrochloric acid and zeroed to the top
calibration mark of 0.00 cm3.
- Universal indicator is added to
the flask and the indigestion powder should turn it blue -
alkaline.
- The acid is carefully and slowly
added until the indicator turns green - neutral at the end-point
of the titration.
- You then read the volume of acid
required to neutralise the ant-acid powder.
- The bigger the volume of acid
required for neutralisation, the more effective the indigestion
powder per mass of powder.
- Repeat the procedure with
another brand of indigestion powder using the same standard acid
solution (fair test).
- After this, the method is
essentially the same as described below.
- 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.
- 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.
- Universal indicator, and most
other acid-base indicators, work for strong acid and alkali titrations, but
universal indicator is a 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.
- 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.
- The
Acids, Bases, pH page
section (2) lists common indicators.
- The theory, and examples of
strong/weak acids/alkalis (soluble bases) are described on the
Extra Aqueous
Chemistry page section 3,
- and the
Acids, Bases, pH page
section (7) explains the changes in pH in the titration.
- Advanced level theory of indicators
and titrations and
advanced acid-alkali
titration questions (GCE-AS-A2-IB students only!)

Self-assessment Quiz on titrations [vct]
type in answer
Honly or
multiple choice
Honly
See also
Advanced level
GCE-AS-A2 acid-alkali titration calculation questions
OTHER CALCULATION PAGES
-
What is relative atomic mass?,
relative isotopic mass and calculating relative atomic mass
-
Calculating relative
formula/molecular mass of a compound or element molecule
-
Law of Conservation of Mass and simple reacting mass calculations
-
Composition by percentage mass of elements
in a compound
-
Empirical formula and formula mass of a compound from reacting masses
(easy start, not using moles)
-
Reacting mass ratio calculations of reactants and products
from equations
(NOT using
moles) and brief mention of actual percent % yield and theoretical yield,
atom economy
and formula mass determination
-
Introducing moles: The connection between moles, mass and formula mass - the basis of reacting mole ratio calculations
(relating reacting masses and formula
mass)
-
Using
moles to calculate empirical formula and deduce molecular formula of a compound/molecule
(starting with reacting masses or % composition)
-
Moles and the molar volume of a gas, Avogadro's Law
-
Reacting gas volume
ratios, Avogadro's Law
and Gay-Lussac's Law (ratio of gaseous
reactants-products)
-
Molarity, volumes and solution
concentrations (and diagrams of apparatus)
-
How to
do volumetric titration calculations e.g. acid-alkali titrations
(this page)
-
Electrolysis products calculations (negative cathode and positive anode products)
-
Other calculations
e.g. % purity, % percentage & theoretical yield, volumetric titration
apparatus, dilution of solutions
(and diagrams of apparatus), water of crystallisation, quantity of reactants
required, atom economy
-
Energy transfers in physical/chemical changes,
exothermic/endothermic reactions
-
Gas calculations involving PVT relationships,
Boyle's and Charles Laws
-
Radioactivity & half-life calculations including
dating materials
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