Molarity, moles & mass and Volumetric titration
calculations e.g. acid-alkali titrations
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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.
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You should be able to carry out calculations involving neutralisation reactions in aqueous
solution given the balanced equation or from your own practical results.
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Note again: 1dm3 = 1 litre = 1000ml = 1000 cm3, so dividing cm3/1000 gives dm3.
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and other useful formulae or relationships are:
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moles = molarity (mol/dm3) x volume (dm3 = cm3/1000),
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molarity (mol/dm3) = mol / volume (dm3 = cm3/1000),
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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)
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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.
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Using a suitable indicator it was found that 15.0 cm3 of the acid was required to
neutralise the alkali.
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Calculate the molarity of the sodium hydroxide and its concentration in
g/dm3.
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moles = molarity x volume (in dm3 = cm3/100)
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moles HCl = 0.200 x (15.0/1000) = 0.003 mol
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moles HCl = moles NaOH (1 : 1 in equation)
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so there is 0.003 mol NaOH in 25.0 cm3
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scaling up to 1000 cm3 (1 dm3), there are ...
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0.003 x (1000/25.0) = 0.12 mol NaOH in 1 dm3
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molarity of NaOH is 0.120 mol dm-3 (or 0.12M)
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since mass = moles x formula mass, and Mr(NaOH) = 23 + 16 + 1 =
40
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concentration in g/dm3 = molarity x formula mass
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concentration in g/dm3 is 0.12 x 40 = 4.80 g/dm3
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Example 12.2: Given the equation 2KOH(aq) + H2SO4(aq) ==> K2SO4 + 2H2O(l)
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20.0 cm3 of a sulphuric acid solution was titrated with 0.0500 mol dm-3 (0.05M) potassium
hydroxide.
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If the acid required 36.0 cm3 of the alkali KOH for neutralisation what was the
concentration of the acid?
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moles = molarity x volume (in dm3 = cm3/100)
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mol KOH = 0.0500 x (36.0/1000) = 0.0018 mol
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mol H2SO4 = mol KOH / 2 (because of 2 : 1 ratio in equation above)
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mol H2SO4 = 0.0018/2 = 0.0009 (in 20.0 cm3)
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scaling up to 1000 cm3 of solution = 0.0009 x (1000/20.0) = 0.0450 mol
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mol H2SO4 in 1 dm3 = 0.0450
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so molarity of H2SO4 = 0.0450 mol dm-3 (0.045M)
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since mass = moles x formula mass, and Mr(H2SO4) = 2 + 32 + (4x16) = 98
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concentration in g/dm3 is 0.045 x 98 = 4.41 g/dm3
Example 12.3: