Reacting masses, concentration of solution and volumetric titration
calculations
Concentration of a solution
For quantitative chemistry, it is important to know the concentration of solutions and be able to
do calculations based on experimental results for chemical analysis.
The concentration of a solution is often measured in grams per decimetre
cubed
so the units are expressed as gdm-3, or g/dm3, which was g/litre!).
concentration = mass / volume (c = m / V)
concentration in gdm-3, mass in g, volume in dm3
Its really important remember that 1 dm3 = 1000 cm3 or 1000 ml
and cm3/1000 = dm3
rearrangements of the formula using the triangle to help gives
mass = concentration x volume (m = c x V)
and volume = mass / concentration (V = m / c)
In the example questions I have used the shorthand formulae c = m / V, m = c x V and V = m / c
and RFM as an abbreviation of relative formula mass or molecular mass, and other shorthands:sf means numbers rounded to 2/3/4 significant figures and dp means round to 2/3/4 decimal places
Example 6b.1
(a) What is the concentration of a salt solution if you dissolve 10g of sodium chloride in 250 cm 3 of
water?
250 cm3 is equal to 250/1000 = 0.25dm3
therefore the concentration c = m / V = 10/0.25 = 40 g/dm3
(40g/litre in old units, still in common use!)
(b) What mass of the salt is required to make 200 cm3 of concentration 15g/dm3?
V = 200/1000 = 0.2 dm3
m = c x V = 15 x 0.2 = 3.0 g
(c) If you were given 8.0 g of salt, what volume of water, in dm3 and cm3, should you dissolve it in, to
give a salt solution of concentration of 5g/dm3?
V = m / c = 8 / 5 = 1.6 dm3
V = 1.6 x 1000 = 1600 cm3
Examples 6b.2, 3 & 4 Quantitative calculations involving titrations
The next three examples are more complicated and involve using reacting mass ratio calculations
which have been covered in detail in section
Example 6b.2
0.5g of sodium chloride was dissolved in water and you are given a silver nitrate solution of
concentration 30g/dm3. When you mix the two solutions you get a white precipitate of silver
chloride.
Given the equation and the RFMs calculated from the atomic masses
Na = 23, Cl = 35.5, Ag = 107.8, O = 16, N = 14
sodium chloride
+ silver nitrate
===>
NaCl(aq)
+ AgNO3(aq)
===> AgCl(s)
(a) RFM 58.5
169.8
silver chloride
(143.3)
+ sodium nitrate
+ NaNO3(aq)
(85) (not needed)
Using an accurate burette, precisely what volume (in cm3) of the silver nitrate solution must be
added to the sodium chloride solution in order to precipitate the maximum amount of silver chloride
without wasting a drop of quite a costly solution! Three steps to the calculation:- (a) calculate the
relative formula masses of the reactants, (b) the mass of silver nitrate reacting and (c) the volume of
silver nitrate needed.
(b) From the equation calculate the mass of silver nitrate that reacts with 0.5 g of sodium chloride.
According to the symbol equation one RFM or 'molecule' of sodium chloride reacts with one RFM or
'molecule' of silver nitrate.
NaCl
:
AgNO3
58.5
:
169.8
0.5g
:
xg
Solving the ratio gives x = 0.5 x 169.8 / 58.5 = 1.451g of silver nitrate
(b) From your answer to (a) calculate the volume of silver nitrate needed.
V = m / c = 1.451 / 30 = 0.0484 dm3
V = 1000 x 0.0484 = 48.4 cm3
Example 6b.3
The reaction between hydrochloric acid and sodium hydroxide is
hydrochloric acid
+ sodium hydroxide
===> sodium chloride
+ water
HCl(aq)
+ NaOH(aq)
===> NaCl(aq)
+ H2O(l)
RFM = 36.5
RFM = 40
According to the symbol equation one RFM or 'molecule' of hydrochloric acid reacts with one RFM
or 'molecule' of sodium hydroxide.
(The atomic masses involved are H = 1, Cl = 35.5, Na = 23 and O = 16, check the RFM for yourself)
Therefore from the relative formula masses (RFMs) 36.5g of HCl reacts with 40g of sodium
hydroxide.
You are given a solution of hydrochloric acid of 7.3 g/dm3 (of HCl). 25.0 cm3 of sodium hydroxide
solution was pipetted into a conical flask. On titration with the acid solution using a burette and
suitable indicator, it was found that 14.6 cm3 of the acid solution was required to completely
neutralise the alkaline sodium hydroxide. Calculate the concentration of the sodium hydroxide
solution in g/dm3 via the stages outlined below
(a) The mass of hydrochloric acid reacting.
V = 14.6 cm3/1000 = 0.0146 dm3
m = c x V = 7.3 x 0.0146 = 0.1066 g HCl
(b) Calculate the mass of NaOH that reacts with 0.1066 g of HCl
HCl
: NaOH
36.5
: 40
0.1066g
: xg
Solving the ratio x = 0.1066 x 40/36.5 = 0.1168 g
(c) Calculate the concentration of NaOH
V = 25 cm3/1000 = 0.025 dm3
c = 0.1168/0.025 = 4.67 g/dm3 (2dp, 3sf)
Now it is possible to titrate the acid solution with the alkali solution (or vice versa) to obtain an
unknown concentration of one of the solutions. One concentration must be known and the two
volumes (acid and alkali) which react together exactly i.e. no excess of either reactant solution. How
to do this is described in
Example 6b.4
Citric acid is the most common acid in citrus fruits and can be estimated by titration with sodium
hydroxide.
citric acid
+ sodium hydroxide
===> sodium citrate
+ water
+ 3NaOH(aq)
===>
+ H2O(l)
(aq)
(aq)
RFM = 40
RFM = 192
but reacting mass is
for C6H8O7
3 x 40 = 120
You are a given of sodium hydroxide with a concentration of 12.0 g/dm3.
10 cm3 of squeezed lemon juice was pipetted into a flask. Using a burette and suitable indicator the
lemon juice required a titration volume of 35.5 cm3 of the sodium hydroxide solution to completely
neutralise it.
Assuming all the acidity is due to citric acid, calculate the concentration of citric acid in the lemon
juice.
(a) What mass of sodium hydroxide reacted with the lemon juice?
V = 35.5 cm3/1000 = 0.0355 dm3
m = c x V = 12.0 x 0.0355 = 0.426g NaOH
(b) What mass of citric acid reacted with the 0.426 g of sodium hydroxide?
C6H8O7
: 3NaOH
192
: 3 x 40 = 120
xg
: 0.426 g
Solving the ratio x = 0.426 x 192/120 = 0.6816 g C6H8O7
(c) Calculate the concentration of citric acid in the lemon juice?
V = 10 cm3/1000 = 0.01 dm3
c = m / V = 0.6816/0.01 = 68.2 g/dm3 C6H8O7
QUESTIONS answers at the end so you can selectively printout the questions and return for the
answers!


Question 6b.1: A potassium sulfate solution A has a concentration of 6.5g/dm3
o
(a) What mass of potassium sulfate is present in 25cm3 of solution A?
o
(b) What volume of solution A, in cm3, contains 3.0g of potassium sulphate?
o
(c) What mass of the salt is required to make 250 cm3 of a solution B so it contains
3.75 g/dm3 of potassium sulfate?
o
(d) If 12.5g of potassium sulphate was dissolved in 1500 cm3 of water to make up
solution C, what is the concentration of the salt?
Question 6b.2: Nitric acid - sodium hydroxide titration
o
The reaction equation for the neutralisation of nitric acid by sodium hydroxide to
form sodium nitrate is
o
HNO3(aq) + NaOH(aq) ===> NaNO3(aq) + H2O(l)
o
(a) Calculate the relative formula masses of nitric acid, sodium hydroxide and
sodium nitrate

o
Atomic masses: H = 1, N = 14, O = 16, Na = 23
25.0 cm3 of a nitric acid solution was pipetted into a conical flask and titrated with a
sodium hydroxide solution of concentration 5.0g/dm3. If it took 16.8 cm3 of the alkali
to neutralise the acid, calculate the following

o
(b) The mass of sodium hydroxide that reacted with the nitric acid.
o
(c) Calculate the mass of nitric acid that reacted with the sodium hydroxide.
o
(d) What was the concentration of the nitric acid in g/dm3?
o
(e) If the resulting neutral solution was carefully evaporated to dryness, what mass
of sodium nitrate salt crystals would be left as a residue?
Question 6b.3: Limewater analysis
o
The symbol equation for the neutralisation reaction between calcium hydroxide
solution (limewater) and hydrochloric acid to form calcium chloride and water is
o
Ca(OH)2(aq) + 2HCl(aq) ===> CaCl2(aq) + 2H2O(l)
o
50 cm3 of a limewater solution was completely neutralised by 9.7 cm3 of a
hydrochloric acid solution. If the concentration of the hydrochloric acid was 3.65 g
HCl/dm3 calculate the concentration of calcium hydroxide in the water by the
following method.
o
(a) calculate the relative formula masses of the reactants.


Atomic masses: Ca = 40, O = 16, H = 1, Cl = 35.5
o
(b) What mass of hydrochloric acid (as HCl) reacted with the limewater?
o
(c) What mass of calcium hydroxide reacted with the hydrochloric acid?
o
(d) What is the concentration of calcium hydroxide in g/dm3?
Question 6b.4: Sulfuric acid - potassium hydroxide reaction to make potassium sulfate
o
The neutralisation reaction between sulphuric acid and potassium hydroxide is
o
H2SO4(aq) + 2KOH(aq) ===> K2SO4(aq) + 2H2O(l)
o
A solution of potassium hydroxide contained 100g/dm3.
o
A solution of sulfuric acid contained 120g/dm3.
o
What volume in cm3, of the sulfuric acid solution, is required to neutralise 250 cm3
of the potassium hydroxide solution?
o
Atomic masses: H = 1, S = 32, O = 16, K = 39
ANSWERS

Question 6b.1: A potassium sulfate solution has a concentration of 6.5g/dm3
o
o
o
o

(a) What mass of potassium sulfate is present in 25cm3 of the solution?

25cm3 = 25/1000 = 0.025 dm3

m=cxV

mass = 6.5 x 0.025 = 0.1625 g
(b) What volume of solution, in cm3, contains 3.0g of potassium sulphate?

V = m / c = 3.0 / 6.5 = 0.462 dm3 (to 3 sf)

V = 1000 x 0.462 = 462 cm3
(c) What mass of the salt is required to make 250 cm3 of a solution B so it contains
3.75 g/dm3 of potassium sulfate?

volume = 250/1000 = 0.25 dm3

m = c x V = 3.75 x 0.25 = 0.9375 g
(d) If 12.5g of potassium sulphate was dissolved in 1500 cm3 of water to make up
solution C, what is the concentration of the salt?

V = 1500 cm3/1000 = 1.5 dm3

c = m / V = 12.5/1.5 = 8.33 g/dm3
Question 6b.2: Nitric acid - sodium hydroxide titration
o
(a)
nitric acid
+ sodium hydroxide
===>
sodium nitrate
+ water
HNO3(aq)
+ NaOH(aq)
===>
NaNO3(aq)
+ 2H2O(l)
RFM 63
40
85
-
o
25.0 cm3 of a nitric acid solution was pipetted into a conical flask and titrated with a
sodium hydroxide solution of concentration 5.0g/dm3. If it took 16.8 cm3 of the alkali
to neutralise the acid, calculate the following
o
(b) The mass of sodium hydroxide that reacted with the nitric acid.
o

V = 16.8 cm3/1000 = 0.0168 dm3

m = c x V = 5.0 x 0.0168 = 0.084g NaOH
(c) Calculate the mass of nitric acid that reacted with the sodium hydroxide.


o
o
HNO3
: NaOH
63
: 40
xg
: 0.084 g
Solving the ratio: x = 0.084 x 63/40 = 0.1323 g HNO3
(d) What was the concentration of the nitric acid in g/dm3?

V = 25.0 cm3/1000 = 0.025 dm3

c = m / V = 0.1323 / 0.025 = 5.29 g/dm3 (3sf, 2dp)
(e) If the resulting neutral solution was carefully evaporated to dryness, what mass
of sodium nitrate salt crystals would be left as a residue?


NaOH
: NaNO3
40
: 85
0.084 g
: xg

Solving the ratio: x = 0.084 x 85/40 = 0.1785 g NaNO3

You can also use HNO3 instead of NaOH i.e. 63 : 85, it doesn't matter which
you use.
Question 6b.3: Limewater analysis
o
The symbol equation for the neutralisation reaction between calcium hydroxide
solution (limewater) and hydrochloric acid to form calcium chloride and water is
o
50 cm3 of a limewater solution was completely neutralised by 9.7 cm3 of a
hydrochloric acid solution. If the concentration of the hydrochloric acid was 3.65 g
HCl/dm3 calculate the concentration of calcium hydroxide in the water by the
following method.
o
(a) Calculate the relative formula masses of the reactants.
o
calcium hydroxide
+ hydrochloric acid
===>
Ca(OH)2(aq)
+ 2HCl(aq)
===> CaCl2(aq)
RFM 74
o
36.5
calcium chloride
-
(b) What mass of hydrochloric acid (as HCl) reacted with the limewater?
+ water
+ 2H2O(l)
o

V = 9.7 cm3/1000 = 0.0097 dm3

m = c x V = 3.65 x 0.0097 = 0.0354 g HCl
(c) What mass of calcium hydroxide reacted with the hydrochloric acid?


o

Ca(OH)2
: 2HCl
74
: 2 x 36.5 = 73
xg
: 0.0354 g
Solving the ratio: x = 0.0354 x 74/73 = 0.03588 g Ca(OH)2
(d) What is the concentration of calcium hydroxide in g/dm3?

V = 50 cm3/1000 = 0.05 dm3

c = m / V = 0.03588/0.05 = 0.718 g/dm3 Ca(OH)2
Question 6b.4: Sulfuric acid - potassium hydroxide reaction to make potassium sulfate
o
H2SO4(aq) + 2KOH(aq) ===> K2SO4(aq) + 2H2O(l)
o
Potassium hydroxide contained 100g/dm3 and sulfuric acid contained 120g/dm3.
o
What volume in cm3, of the sulfuric acid solution, is required to neutralise 250 cm3
of the potassium hydroxide solution?
o
To calculate mass of KOH
o

V = 250 cm3/1000 = 0.25 dm3

m = c x V = 100 x 0.25 = 25g of KOH
To calculate mass of H2SO4 needed


o
H2SO4
: 2KOH
98
: 2 x 56 = 112
xg
: 25 g
Solving the ratio: x = 25 x 98/112 = 21.875 g H2SO4
To calculate the volume of sulphuric acid needed.

V = m / c = 21.875/120 = 0.1823 dm3

0.1823 x 1000 = 182 cm3 (3sf, nearest cm3)