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Determining Calcium Ion Concentration in Water by Complexometric Titration

Key Concepts

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Precipitating the Magnesium Ions.

Calcium and magnesium belong to the same group in the Periodic Table, Group 2.
This means that the chemistry of calcium and magnesium is similar.
We need to choose a reagent to add to the water that forms an insoluble solid (a precipitate) with magnesium but not with calcium.
We can do this by comparing the solubility product, Ksp, values for a number of calcium and magnesium salts as shown in the table below:

Ksp
magnesiumcalcium
carbonate3.5 x 10-82.8 x 10-9
oxalate1.0 x 10-84.0 x 10-9
fluoride6.5 x 10-94.0 x 10-11
hydroxide1.8 x 10-115.0 x 10-6

Magnesium carbonate, oxalate and fluoride are all more soluble than calcium carbonate, oxalate and fluoride.

Magnesium hydroxide, however, is less soluble than calcium hydroxide.

It should be possible to precipitate magnesium hydroxide out of the water without precipitating out all the calcium as calcium hydroxide.

Mg(OH)2 Mg2+(aq) + 2OH-(aq)
Ksp(Mg(OH)2) =[Mg2+][OH-]2
1.8 x 10-11 = [Mg2+][OH-]2
Let y = [Mg2+] at equilibrium
Then [OH-] = 2y at equilibrium
So 1.8 x 10-11 = y x (2y)2
1.8 x 10-11 = y x 4y2
1.8 x 10-11 = 4y3
4.5 x 10-12 = y3
y = 1.65 x 10-4
[Mg2+] = 1.65 x 10-4 mol L-1
[OH-] = 2 x 1.65 x 10-4
= 3.30 x 10-4 mol L-1
Ca(OH)2 Ca2+(aq) + 2OH-(aq)
Ksp(Ca(OH)2) =[Ca2+][OH-]2
5.0 x 10-6 = [Ca2+][OH-]2
Let y = [Ca2+] at equilibrium
Then [OH-] = 2y at equilibrium
So 5.0 x 10-6 = y x (2y)2
5.0 x 10-6 = y x 4y2
5.0 x 10-6 = 4y3
1.25 x 10-6 = y3
y = 1.08 x 10-2
[Ca2+] = 1.08 x 10-2 mol L-1
[OH-] = 2 x 1.08 x 10-2
= 2.16 x 10-2 mol L-1
Under the same conditions, the concentration of hydroxide ions in the calcium hydroxide solution is about 100 times more than the concentration of hydroxide ions in the magnesium hydroxide solution.
Calcium hydroxide is more soluble than magnesium hydroxide under the same conditions.

An aqueous solution of sodium hydroxide can be used to precipitate out the magnesium hydroxide.

Assume that hard water contains a maximum 1 g L-1 Mg2+ and 1 g L-1 Ca2+
To precipitate out Mg(OH)2, its ion product, Q,
must be greater than its solubility product, Ksp.
[Mg2+] = moles ÷ volume (L)
[Mg2+] = 1/24.31 ÷ 1 = 0.041 mol L-1
Q = [Mg2+][OH-]2 > Ksp
[Mg2+][OH-]2 > 1.8 x 10-11
[0.041][OH-]2 > 1.8 x 10-11
[OH-]2 > 4.4 x 10-10 mol L-1
[OH-] > 2.1 x 10-5 mol L-1
To precipitate out Ca(OH)2, its ion product, Q,
must be greater than its solubility product, Ksp.
[Ca2+] = moles ÷ volume (L)
[Ca2+] = 1/40.08 ÷ 1 = 0.025 mol L-1
Q = [Ca2+][OH-]2 > Ksp
[Ca2+][OH-]2 > 5.0 x 10-6
[0.025][OH-]2 > 5.0 x 10-6
[OH-]2 > 2.0 x 10-4 mol L-1
[OH-] > 1.4 x 10-2 mol L-1
In order to precipitate out Mg(OH)2 the [OH-] must be greater than 2.1 x 10-5 mol L-1,
but [OH-] must be less than 1.4 x 10-2 so that the Ca(OH)2 does NOT precipitate out.
The pH of the water sample should be adjusted to be about 12, at 25oC,
using NaOH(aq) to precipitate out Mg(OH)2 without precipitating out Ca(OH)2.
 

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The Titration

After the magnesium ions have been precipitated out of the hard water by the addition of NaOH(aq) to form white Mg(OH)2(s), the remaining Ca2+ ions in solution are titrated with EDTA solution.

The Calculations

The balanced chemical equation for the reaction between Ca2+ and EDTA is shown below:
Ca2+ + H2EDTA2- → CaEDTA2- + 2H+

The stoichiometric (mole) ratio of Ca2+ : EDTA is 1 : 1

That is, at the equivalence point, the moles of EDTA added to the water equals the moles of Ca2+ originally present in the water
At the equivalence point: moles Ca2+ = moles EDTA

moles of EDTA = EDTA concentration (mol L-1) x volume EDTA used (L)

volume used = average of the concordant titres (L)

moles Ca2+ in hard water sample = moles EDTA used = EDTA concentration (mol L-1) x volume EDTA used (L)

concentration (molarity) of Ca2+ in hard water = moles Ca2+ ÷ volume of water sample used (L)

[Ca2+] = (EDTA concentration (mol L-1) x volume EDTA used (L)) ÷ volume water sample (L)

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Worked Example

A student has been asked to determine the concentration of calcium ions, in mg L-1, in a 1 L sample of water.
The student places 50.0 mL of the water sample in a conical flask.
NaOH(aq) is added to the sample until the pH is about 12. The solution is swirled for a couple of minutes to completely precipitate all the magnesium ions as white Mg(OH)2(s).
A drop of Patton and Reeder's Indicator is added to the water sample. The water sample changes colour to red.
The burette is filled with 50.00 mL of 0.010 mol L-1 EDTA solution.
The student conducts a rough titration and discovers the endpoint of the titration, when the indicator changes colour to blue, occurs when about 9 mL of EDTA have been added to the water.
The student then conducts 3 more titrations, taking great care.
The results of these titrations are shown below.

Titration no.volume of EDTA (mL)
18.76 mL
28.80 mL
38.78 mL

Calculate the concentration of calcium ions in the hard water sample.

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