Comparing the Concentration of Some Alkalis in Saturated Solutions

When metals from groups one and two of the periodic table combine with an OH molecule, the type of intramolecular bonding is ionic bonding. The metals lose one or more electrons and so become positively charged. The non-metal ions/ molecules gain electrons and so become negatively charged. The oppositely charged ions attract each other to form a rigid 3-D lattice a. Each ion in the lattice is surrounded by others of opposite charge e.g. NaOH:

Due to their polar nature, ionic compounds are usually soluble in polar solvents, for example water:

Water is polar due the difference in electronegativity between the oxygen and hydrogen atoms.

This is represented by ?+ and ?-.

When an ionic substance dissolves in a solvent, the lattice must be broken and the separated species must then be surrounded by individual solvent molecules b. The solubility of a solute in water, at a given temperature is the maximum amount of it that will dissolve in 100 grams of water at that temperature.

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When this amount is reached, it gives a saturated solution c. In a saturated solution, e.g. Ca(OH)2 (aq), Ca2+ and OH- ions are in equilibrium with the solid calcium hydroxide:

Ca(OH)2 (aq) Ca2+ + 2OH-

The solubility of group one and two hydroxides varies greatly, however, clear trends can be seen:

Metal Hydroxide

Mr (g)

Moles sat (mol/100g)

LiOH

023.9

5.16×10-1

NaOH

040.0

1.05

KOH

056.1

1.71

RbOH

102.5

1.69(303K)

CsOH

150.0

2.02(303K)

FrOH

Be(OH)2

026.0

Mg(OH)2

058.3

2.00×10-5

Ca(OH)2

074.1

1.53×10-3

Sr(OH)2

121.6

3.37×10-3

Ba(OH)2

171.0

1.50×10-2

Ra(OH)2

(303K) = at a temperature of 303K or 30oC

Clearly, as you move down the group, the solubility increases and solubility decreases as you move across a period.

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This is due to the size of ions and their reactivity.

As you move down groups one and two, the atomic radii of the atoms increases. This is because:

1. The number of electron shells is increasing;

2. Distance of outer electron from nucleus increases;

3. Increases shielding of positive nuclear charge ?outer electrons held less tightly;

4. Increased electron repulsion.

The increases size of metal ions at the base of groups one and two means that the distance between OH- and X+/2+ ions will be greater, therefore the electrostatic attraction between them is less. Subsequently the solid 3-D lattice will break apart much more easily and the hydroxide will be more soluble than those with smaller metal ions b.

As you move across a period, the solubility decreases. This is due to a decreased atomic radius and an increased charge on the ions. With a smaller atomic radii, the distance between OH- and metal ions decreases, thus the electrostatic attraction is much stronger. Therefore the lattice will not break apart as easily, as more energy will be needed for this to occur ?solubility will be lower.

The greater the charge on the ion, the greater the attraction between them and the more energy that will be required to break apart the lattice. That is, the attraction of Mg2+ – OH- ? Na2+ – OH- ? more energy is required to separate the ionic bonds of the lattice, thus solubility decreases.

Finally there is the difficulty of ionic bonds with covalent character – i.e. in the case of Be(OH)2, which is insoluble in water. The ionic bond is polarised and subsequently the compound has covalent character, thus it is insoluble in the polar H2O.

Apparatus For Experiment

* Burette

* White tile

* Ca(OH)2

* KOH (aq)

* NaOH (aq)

* LiOH (aq)

* Pipette and pipette filler

* Phenolphthalein indicator

* Distilled H2O

* Volumetric flasks

* Beaker

* Conical flasks (250 cm3 and 100 cm3)

* 0.05 moldm-3 HCl

* 0.10 moldm-3 HCl

* 0.20 moldm-3 HCl

Diagram of Apparatus

Method

In order to determine the concentration of the alkali hydroxides, it is necessary to titrate them against HCl. Hydrochloric acid is a strong acid, i.e. it fully dissociates in H2O: HCl goes to H+ and Cl-. The hydroxides can also be classed as strong, as they too fully dissociate. Thus it is appropriate to use phenolphthalein indicator, which is colourless in acidic solutions and pink in basic ones.

From the table on page one, it is possible to calculate the concentrations of the hydroxides that are to be used in this experiment; purely by multiplying by ten (mol/1000g = mol/dm-3). These were chosen because of their suitable concentrations. The chosen hydroxides are Ca(OH)2, LiOH, NaOH and KOH:

Hydroxide

Moles sat (mol/100g)

Concentration (mol/1000g)

Ca(OH)2

1.53×10-3

1.53×10-2

LiOH

5.16×10-1

5.16

NaOH

1.05

10.5

KOH

1.71

17.1

To Calculate Concentration of Ca(OH)2 solution:

1. 0.0153 moldm-3 is suitable for concentration. Pipette 50 cm3 of Ca(OH)2 (aq) into a 250 cm3 conical flask as this volume will ensure that spillages are avoided. Add four drops of phenolphthalein.

2. Calculate mole of Ca(OH)2 = conc. x vol

= 0.0152 x 0.06 = 7.65×10-4

3. Calculate suitable conc. of HCl to titrate with( approx 15 – 30 cm3 is a suitable titre)

Ca(OH)2 + 2HCl CaCl2 + 2H2O

1 : 2

Moles = 7.65×10-4 : 1.53×10-3

Conc. = Moles / Volume

= 1.53×10-3 / (30/1000) = 0.051

Thus a suitable concentration of HCl to utilise is 0.05 moldm-3.

4. Using a burette, titrate 0.05 moldm-3 HCl against the Ca(OH)2. Titrate until solution reaches the neutralisation point, i.e. just when the pink colour disappears, and the solution becomes colourless. Record the result and repeat the experiment until two readings are obtained which are within 0.1 cm3 of each other.

5. Calculate, using the results of the titration, the experimental value for the conc. of the saturated Ca(OH)2.

Calculating the concentration of LiOH, NaOH and KOH

LiOH + HCl LiCl + H2O

NaOH + HCl NaCl + H2O

KOH + HCl KCl + H2O

Ratios: 1 : 1 : 1 : 1

1. Dilute the solutions as directed in the table below:

Stage of Dilution

LiOH

NaOH

KOH

1

Pipette 10cm3 into a 250cm3 volumetric flask.

Pipette 25cm3 into a 250 cm3 volumetric flask.

Pipette 25cm3 into a 250 cm3 volumetric flask.

2

Make up to 250cm3 with distilled water.

Make up to 250cm3 with distilled water.

Make up to 250cm3 with distilled water.

3

Shake 40 times to ensure solution is uniform.

Shake 40 times to ensure solution is uniform.

Shake 40 times to ensure solution is uniform.

4

Pipette 25cm3 of diluted solution into a new volumetric flask.

Pipette 25cm3 of diluted solution into a new volumetric flask.

5

Make up to 250cm3 with distilled water.

Make up to 250cm3 with distilled water.

6

Shake 40 times to ensure solution is uniform.

Shake 40 times to ensure solution is uniform.

2. Calculate the concentrations of each of the diluted solutions:

Solution

Concentration

LiOH

Moles = conc. x vol

= 5.16 x 0.01 = 0.0516 moles in 250cm3

? in 1 dm3 = 0.0516 x 4 = 0.2064 moldm-3

NaOH

Moles = conc. x vol

= 10.5 x 0.025 = 0.2625 moles in 250cm3

? in 1 dm3 = 0.2625 x 4 =1.05 moldm-3

Moles = conc. x vol

= 1.05 x 0.025 = 0.02625 moles in 250cm3

? in 1dm3 = 0.02625 x 4 = 0.105 moldm-3

KOH

Moles = conc. x vol

= 17.1 x 0.025 = 0.4275 moles in 250cm3

? in 1 dm3 = 0.4275 x 4 =1.71 moldm-3

Moles = conc. x vol

= 1.71 x 0.025 = 0.04275 moles in 250cm3

? in 1dm3 = 0.04275 x 4 = 0.171 moldm-3

3. Select a suitable concentration of HCl to titrate with for each solution (based on 1:1 ratio):

Diluted Solution

Diluted Concentration (moldm-3)

HCl Concentration

(moldm-3)

LiOH

0.2064

0.2

NaOH

0.105

0.1

KOH

0.171

0.2

4. Titrate the selected concentrations of HCl against the diluted solutions (ensuring indicator is added before hand) and record the titre value at the point of neutralisation. Repeat the experiments as before.

5. Calculate the experimental concentrations of each saturated solution.

In order to determine which is the most concentrated of the alkali hydroxides utilised in this experiment, it is necessary to use stoichiometry and calculations to calculate the experimental values of the concentration of the saturated solutions. I predict, based on scientific knowledge and background information, that the most soluble of the four in this experiment will be KOH.

Possible Errors

There are several major errors that must be avoided in this experiment, they are listed with ways to prevent them from occurring:

* When diluting, read the 250cm3 mark at the bottom of the meniscus. Use a fine pipette to add distilled water drop-wise for accuracy if necessary.

* Do not use measuring cylinders as they are inaccurate compared to pipettes.

* Ensure contents of volumetric flask are mixed to form uniform solutions.

* Readings must be made to 0.05cm3 when using a burette.

* Utilise a white tile to ensure that the end point is easily recognised.

* Make sure all experiments are repeated until two recordings are within 0.1 cm3 of each other. This increases the reliability of the results.

* Finally, it is vital that the room temperature is maintained at a constant level. If temperature is raised, then energy is used to aid the breaking of the ionic lattice and subsequently the substance becomes more soluble. Similarly, if temperature is lower, the solubility will decrease.

Safety and Risk Assessment:

According to Cleapss Hazcards, solutions of KOH, NaOH and LiOH ? 0.5 molar are to be labelled corrosive. Eye protection should be worn and gloves also. Ca(OH)2 is not given a hazard classification, but it can be irritating to the eyes ? eye protection should be worn. HCl is not dangerous with the concentrations utilised in this experiment, but eye protection should be worn. The final protection is to use a pipette filler rather than oral suction for pipetting, as the solutions utilised are very corrosive and unsafe.

Bibliography

Source

Author

Page Number/s

Superscript Indicator

Advanced Chemistry

Michael Clugston

Rosalind Flemming

58-59

154-155

a

b

GCSE Chemistry

Rosie Marie Gallagher

Paul Ingram

18

c

Cleapss Hazcards

Cleapss

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Comparing the Concentration of Some Alkalis in Saturated Solutions. (2020, Jun 01). Retrieved from https://studymoose.com/comparing-concentration-alkalis-saturated-solutions-new-essay

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