3.4 Methods Used 86
3.4.3 Interviews and the ‘practice of talking to people’ 104
p01-1= - log [0111 for water pOH = 7
pH = pOH = 7 for water at 25°C.
and pH + p01-1 =
14 at the same temperature.When an acid dissolves in water the (H1 increases and pH <7 but when a base dissolves [OH -]
increases pOH <7 or pH > 7
On the pH scale therefore, acid solutions have pH <7 while alkaline solutions have pH > 7
7
acidic pH alkaline
128
Example:
Calculate the pH of 0.001 HC/ solution HCI is a strong acid
HC_L)--> 1-1*( 4Ct(aq) [H1 = .001mol dm-3
pH = – log H.
= – log .001
=3
Now calculate pH for .005 H 2SO4(aq) 633 Hydrolysis of salts
A solution of an acid will turn blue litmus red while alkaline solution turns red litmus blue. Some salt solutions are acidic or alkaline to litmus. This is because of hydrolysis of the salts. For example, a solution of sodium ethanoate turns red litmus blue and a solution of ammonium chloride turns blue litmus red. The above observation is explained by a reversible reaction occurring between the salts and water which leads to alkaline or acidic behaviour. CH 3 COONat.) +1120(1 ...„—sCH3 COOHN) + NaOHao. Because sodium hydroxide is a strong base and CH 3C0011 is a weak acid, the solution is alkali i.e. the alkali has an upper hand.
for NH4 Cl
NH4 Claq) + H2 0(1) ,--sNF14 0H(.0 + HCI(aq)
HCI is a strong acid but N11 4 01-1 is a weak alkaline (base). The acid has the upper hand and the resulting solution is acidic to litmus. The concept of hydrolysis is very helpful in explaining such observations as above and also in making proper choice of indicator for acid-base titrations.
63.4 Buffer Solutions
Buffer solutions are solutions of a weak acid or base and one of its salts.
These solutions possess the ability to resist pH change when small amounts of acid or base are added to them. Examples of buffers are ethanoic acid and its sodium salt and ammonia solution and ammonium chloride. The first example is acid buffer and the second is an alkaline buffer. The capacity of buffer solutions to resist change of pH is explained by reversible reactions in the acid-salt mixture.
Take for example the buffer with ethanoic acid and its sodium salt. The following reactions occur.
CH3 000H(aq) CH3 C00-(aq)(aq) weak acid
CH3 COONa(aq) --> CH3 C 00-(aq) Na* (m) highly dissociated
When a small amount of H+ is added to the buffer solution the weak acid equilibrium is reversed and un-dissociated CH3 COOH is formed. This will have little effect on the CH 3C00- in solution because of CH3COO-from the second reaction. Similarly when OH - is added the reaction OH- +
Ft --->
H2 O occurs (H*) is reduced and the acid equilibrium shifts in the forward direction. This produces more CH 3 C00- but not enough to make much difference because of the large concentration of CH 3 C00- from the salt dissociation.Buffer solutions are important in reactions that occur only at constant hydrogen ion concentration (pH).
They are very important in the body because most of body metabolism occurs only at constant pH.
63.5 Common ion effect
A saturated solution is one in which no more of solute will dissolve in the solvent at that temperature. A saturated solution is in equilibrium with excess solid in solution.
Take the example of a saturated calcium chloride solution
CaC12(s) Ca1'
(aq) + 20-(aq)
129
If to this solution is added some sodium chloride crystals (Nod), the NaC1 also furnishes C1 - in solution NaC1 Na (N) + CC)
Cl- is the common ion in this example. The additional Ct ions in solutions will shift the CaCl 2 equilibrium in the reverse direction. More solid will come out i.e solubility of CaC1 2 decreases:
The common ion effect can be defined as the decreasein
solubility caused by the presence of an ion ofa salt that is added to the solution of the saltfrom an external source.An industrial application of common ion effect is in the soap industry. Soap is produced by treating oil or fat with sodium hydroxide. Fat or oil + NaOH —>Soap + Propane 1,2,3, trio!.
'Soap is the sodium salt of a fatty acid. Soap is fairly soluble producing Ne and the anion of the fatty acid.
Addition of sodium chloride (NaC1) to the soap solutions will introduce Ne as common ion. The effect is that the solubility of the soap is reduced and more solid soap precipitates out. The yield of soap is by this method increased. This is called salting out of soap. It is a practical application of common ion effect 63.6 Solubility product
For sparingly soluble salts the solubility product is an index of solubility. Salts with high value of solubility product are more soluble than others with low solubility product values.
Take the example of a saturated AgC1 solution.
AgC1(,) AgRato + k =
[Agil [CY][AgCI]
[AgCI] = [Age] [CY] = k g,
k is the solubility product of AgCI. The importance of the solubility ion concept lies in its bearing upon precipitation from solution of metal ions.
The solubility product of AgC1 at 25°C is 1.1 x 10 -70 mol2din-6. In any aqueous solution the product of the molar concentration of Ag' and CY in solution cannot exceed this value. If for silver chloride the product of the concentrations is made to exceed this value. Adding a salt with a common ion or mixing solutions with high ion concentrations of
Aeand CY ions will result in precipitation of the salt.
Example
The solubility product of AgCI is 1.1 x 101 ° at 25°C. Explain why precipitation occurs when a drop of 0.01moldm-3 NaC1 is added to 10.0cm 3 of 0.l0moldm-3 Ag NO3 solution in a test tube.
Answer:
AgNO3 —> Ag+00 + NO3 tarn Dissociation is assumed complete [Ag] = .10
[NO3] = .10
For AgCI to be precipitated [Ag] [CY] > 1.1 x 10-10
[CY] >
1.1 x 10-10= 1.1 x 10 9 0.1
The amount of CY is so small that a tiny drop of 0.01 molder' sodium chloride solution will give precipitate.
You will recall that this is a test for chlorides.
130
6.4 Conclusion
Reversible reactions are many. This unit is a demonstration of the equilibrium law and Le Chattelier's principle. When applied to industrial processes, reaction conditions are chosen to get optimum yield of products. The application of the equilibrium law to aqueous equilibrium has led to developments of many concepts like acid dissociation constant, common ion effect, solubility product and the pH scale. These concepts are used to explain and quantify some observations in chemical analysis.
6.5 Summary
• The Haber and contact processes are discussed.
• Le Chatelier's principle is used to justify the reaction conditions for optimum yield of product.
• Applications of the equilibrium law and Le Chatelier's principle to aqueous equilibria are enumerated and explained.
6.6 Tutor-Marked Assignments
1. The following reaction conditions are used in the Haber process.
(i) Pressure 200 atmospheres (ii) Temp 450°C
(iii) Catalyst Iron
Justify the above reaction conditions. (Ammonia production is from I-12 and N2 and the process is exothermic)
2. Explain the following observations
(a) An aqueous solution of FeCI3 is acidic to litmus but a solution of Na 2 CO (b) Explain the terms.
(i) Salting out
(ii) Common ion effect (iii) Solubility product
alkaline to litmus.
6.7 References
Rajah, S.T., Teibo, B.O., Onwu G. and Obikwere A., (2002). Senior Secondary Chemistry Textbook 2 Lagos, Longman Publishers.
Osei Yaw Ababio, (2002). New School Chemsny Onitsha, Africans — FEP Publishers.
131