Titration calculations
- Work out the moles of acid and base at the
start
- Work out the excess moles of acid or base (the
rest will be neutralised)
- Work out the new [H+] or new [OH-]
and then the pH
Calculate
the pH of the solution formed when 20 cm3 of 0.10 mol dm-3
HCl is added to 30 cm-3 of 0.04 mol dm-3 NaOH
Moles HCl = 0.10 x 20/1000
= 0.0020 mol
Moles NaOH = 0.04 x 30/1000
= 0.0012 mol
Excess HCl = 0.0020 – 0.0012
= 0.0008 mol
Total
volume = 50 cm3
New [H+] = moles/volume
= 0.0008/(50/1000)
= 0.016 mol dm-3
pH = -log(0.016)
= 1.80
4) Buffer
solutions
- A buffer solution minimises pH changes on
addition of an acid or base
- Buffer solutions are important for controlling
pH in blood (so that enzymes are not denatured) and shampoos (so that eyes
do not sting and skin is not damaged)
a) Acidic buffers
- An acidic buffer consists of a weak acid and
the salt of a weak acid (e.g. ethanoic acid & sodium ethanoate)
- For ethanoic acid/sodium ethanoate, the
following equilibrium exists:
CH3COOH(aq) ¾ CH3COO-(aq) +
H+(aq)
- If a small amount of acid is added, equilibrium
will shift to the left to remove the added H+. The following
reaction occurs: CH3COO- + H+ ® CH3COOH
- If a small amount of base is added, the OH-
will react with H+ to form water. The equilibrium will shift to
the right to replace the H+ that has been removed. The
following reaction occurs: CH3COOH ® CH3COO- + H+
b) Calculating
the pH of a buffer solution
- The pH of an acidic buffer can be calculated
using the Ka expression for the weak acid e.g. for ethanoic
acid/sodium ethanoate:
Ka = [CH3COO-][H+]
[CH3COOH]
Rearranging gives:
[H+]
= Ka x [CH3COOH]
[CH3COO-]
Calculate
the pH of a buffer solution containing equal volumes of 2.5 mol dm-3
HCOONa and 1.0 mol dm-3 HCOOH (Ka = 1.6 x 10-4
mol dm-3)
[H+] = Ka x [HCOOH]/[HCOONa]
= 1.6 x 10-4 x
1.0/2.5
= 6.4 x 10-5 mol dm-3
pH =
4.19
Source: OCR June 2003
paper
- The pH of a particular buffer depends on the
value of Ka and the ratio of [CH3COOH]
to [CH3COO-]
d) pH of Blood
- The
pH of blood is kept in the range 7.35-7.45
- This
is achieved by dissolved carbon dioxide acting as a buffer solution
- The
relevant equations are:
CO2 + H2O ¾ H2CO3
H2CO3 ¾ HCO3- + H+
- H2CO3
is carbonic acid. HCO3- is the hydrogencarbonate ion
- If
the blood pH is less than 7.35 (too acidic), more CO2 is breathed
out. Both equilibria shift to the left to produce more CO2 and
[H+] is reduced
- If
the blood pH is more than 7.45 (too alkaline), less CO2 is
breathed out. Both equilibria shift to the right to remove CO2
and [H+] is increased
5) Enthalpy
of neutralisation
- Enthalpy
of neutralisation is defined as the
change in enthalpy
that occurs when an acid and base undergo a neutralisation reaction to form one
mole of water i.e.
H+(aq)
+ OH-(aq) ® H2O(l)
- This
is a calorimetry calculation e.g.
Example
50 cm3 of 1.0 mol dm-3
hydrochloric acid was added to 50 cm3 of 1.0 mol dm-3
sodium hydroxide solution. The
temperature rose by 6.8°C. Calculate the enthalpy of neutralisation for
this reaction. Assume that the density
of the solution is 1.00 g cm-3, the specific heat capacity of the
solution is 4.18 J g-1 K-1.
HCl + NaOH ® NaCl + H2O
m =
mass of solution = total of acid and alkali = 100g
q =
-mcDT/1000
=
-100 x 4.18 x 6.8/1000
=
-2.8424 kJ
n =
moles HCl = moles NaOH = 1.0 x 50/1000 = 0.05 mol
DH = q/n
=
-2.8424/0.05
=
-56.8 kJ mol-1
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