Why do we do it?
- To determine the concentration of solutions
- To determine the exact volumes of two solutions required to react completely
Titrations can be between:
- AN ACID AND ALKALI SOLUTION- The END POINT ( point at which the reaction is complete) is determined by the colour change of an indicator or thermometrically.
- AN OXIDISING AGENT AND A REDUCING AGENT- the end point is determined by a sudden colour change in one of the reactants.
Titration results can be used to:
- Prepare a salt of sodium, potassium or ammonium
- Determine the mole ratio in which reactants combine so that the balanced chemical equation may be written
- Determine the basicity of an acid
- Determine the concentration of one of the reactants or products
USING TITRATION TO DETERMINE THE MOLE RATIO IN WHICH REACTANTS COMBINE
Let's work an example:25cm3 of KOH solution of concentration 0.8 mol/dm3 neutralise 20 cm3 of an acid H2X containing 0.5 mol/dm3.
(a) Determine the mole ratio in which KOH and H2X combine
(b) Write an equation for the reaction
Answer:
(a) Find the # moles of KOH used:
We know : 1000cm3 of KOH solution contains 0.8 moles
25 cm3 of KOH solution contains 0.02 moles
We can also calculate the number of moles of H2X used:
1000cm3 of H2X solution contains 0.5 moles
20cm3 of H2X solution contains 0.01 moles
Therefore, we know 0.02 moles of KOH reacts with 0.01 moles H2X
Thus the mole ratio is 2 moles of KOH : 1 mole of H2X
(b) 2KOH(aq) + H2X (aq) --> K2X (aq) + 2H2O(l)
USING TITRATION TO DETERMINE THE BASICITY OF AN ACID
Let's work an example: 25cm3 of NaOH solution of a concentration 0.6 mol/dm3 reacts with 37.50cm3 of an acid HnA of concentration of 0.4 mol/dm3. Determine the value of n.
Answer: Number of moles of NaOH used:
If 1000cm3 of NaOH solution contains 0.6 moles
Then 25cm3 of NaOH solution contains 0.015 moles
Number of moles of HnA used:
1000cm3 of HnA solution contains 0.4 moles
37.5 cm3 of HnA solution contains 0.015 moles
So, we can say 0.015 moles of NaOH reacts with 0.015 moles of HnA
Therefore, the mole ratio is 1:1
Ionic equation for the reaction of any acid with any alkali is H+(aq) + OH-(aq) --> H2O(l)
So, HnA must contain 1 mole of H+ ions. Therfore, n=1
Answer: Number of moles of NaOH used:
If 1000cm3 of NaOH solution contains 0.6 moles
Then 25cm3 of NaOH solution contains 0.015 moles
Number of moles of HnA used:
1000cm3 of HnA solution contains 0.4 moles
37.5 cm3 of HnA solution contains 0.015 moles
So, we can say 0.015 moles of NaOH reacts with 0.015 moles of HnA
Therefore, the mole ratio is 1:1
Ionic equation for the reaction of any acid with any alkali is H+(aq) + OH-(aq) --> H2O(l)
So, HnA must contain 1 mole of H+ ions. Therfore, n=1
USING TITRATION TO DETERMINE THE CONCENTRATION OF A REACTANT/PRODUCT
Let's work out an example: 25cm3 of a solution of sodium carbonate required 22cm3 of 0.04M (molar concentration) HCl for complete neutralisation. Calculate the concentration of the sodium carbonate solution in mol/dm3 and g/dm3
Answer:
1) First find the number of moles contained in 22cm3 of HCl.
1000cm3 of HCl solution contains 0.04 moles
Therefore, 22cm3 of HCl solution contains 0.0008 moles
2) Use the chemical equation along with the answer in part1 to determine the number of moles contained in sodium carbonate.
2HCl (aq) + Na2CO3 --> 2NaCl (aq) + H2O(l) + CO2 (g)
From the balanced equation above, 2 moles of HCl react with 1 mole of Na2CO3
Therefore, half the number of moles of the carbonate is needed to react with the acid, which is, 0.00044 moles
3) This number of moles of sodium carbonate (0.00044 moles) is contained in 25cm3
Therefore the molar concentration is:
If 25cm3 of Na2CO3 solution contains 0.00044 moles
Then 1000cm3 of Na2CO3 contains 0.0176 moles MOLAR CONCENTRATION= 0.0176 mol/dm3
4) Now that the molar concentration is known, the mass concentration can be determined
If 1 moles of Na2CO3 has a mass of 106g
Then 0.0176 moles of Na2CO3 has a mass of 1.87g MASS CONCENTRATION= 1.87g/dm3
Answer:
1) First find the number of moles contained in 22cm3 of HCl.
1000cm3 of HCl solution contains 0.04 moles
Therefore, 22cm3 of HCl solution contains 0.0008 moles
2) Use the chemical equation along with the answer in part1 to determine the number of moles contained in sodium carbonate.
2HCl (aq) + Na2CO3 --> 2NaCl (aq) + H2O(l) + CO2 (g)
From the balanced equation above, 2 moles of HCl react with 1 mole of Na2CO3
Therefore, half the number of moles of the carbonate is needed to react with the acid, which is, 0.00044 moles
3) This number of moles of sodium carbonate (0.00044 moles) is contained in 25cm3
Therefore the molar concentration is:
If 25cm3 of Na2CO3 solution contains 0.00044 moles
Then 1000cm3 of Na2CO3 contains 0.0176 moles MOLAR CONCENTRATION= 0.0176 mol/dm3
4) Now that the molar concentration is known, the mass concentration can be determined
If 1 moles of Na2CO3 has a mass of 106g
Then 0.0176 moles of Na2CO3 has a mass of 1.87g MASS CONCENTRATION= 1.87g/dm3
Volumetric Analysis worksheet 1
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Volumetric Analysis worksheet 2
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