Evaluate the Case for and Against Buffer Stock Schemes Essay

Evaluate the Case for and Against Buffer Stock Schemes Essay.

A buffer stock scheme is an intervention carried out by the government which aims to limit fluctuations in the price of a commodity. It involves the government and/or local authorities buying these storage stocks and selling them back to the famer. Price stability is indicated by low inflation whereby the value of money is also stable. A buffer stock is an attempt at stabilising the prices of key commodities. Extract C states that ‘when prices fall governments are more likely to be concerned’ this may be because more people are likely to buy them so the government is more likely to have to buy more from the farmer.

The free market usually determine the prices of commodities such as sugar and tin, yet, without intervention, the prices of coffee and sugar have been unstable as Extract A shows a significant increase in the price of coffee in 2008 and sugar in 2005. Should there be a large rise in supply due to better than expected yields at harvest time, the market supply will shift out – putting downward pressure on the free market equilibrium price.

In this situation, the intervention agency will have to intervene in the market and buy up the surplus stock to prevent the price from falling.

It is easy to see how if the market supply rises faster than demand then the amount of wheat bought into storage will grow. The stable prices help maintain farmers’ incomes and improve the incentive to grow legal crops; this stability enables capital investment in agriculture needed to lift agricultural productivity, as farming has positive externalities it helps to sustain rural communities. The stable prices prevent excess prices for consumers – helping consumer welfare.

However, a minimum price legislation may be applicable as it not only protects the producer, the farmer in this case but it ensures he receives some revenue for his commodities although it can’t be guaranteed as to whether the income the farmer receives will be appropriate and substantial. Extract C also states that ‘very often schemes fail because stocks of commodities such as foodstuffs and metals are more frequently bought than sold’ which means that the government are buying more stocks than are actually being sold, creating a surplus stock.

Extract B says ‘India… would require substantial sugar imports to compensate for the failure of the domestic sugar crop’, another problem with buffer stocks is that the farmer is reliant on good weather, to result in a good harvest and a high yield to cater for the strong demand, and examples such as the ‘Colombian crop’ being ‘damaged by heavy rainfall’ supports the fact that if farmers harvests are affected by natural disasters and unforeseen circumstances, then their job will also be affected as all farmers’ harvests are subject to changing weather conditions.

Furthermore financing the scheme is also difficult as buffer stocks are usually financed by the government, so in effect, the taxpayer, indirectly. As a reult this creates an opportunity cost for the government. It is also difficult to determine how the prices are set as the law of demand and supply suggests that if the price increases, then demand will decrease due to the inverse relationship of the two and the downwards sloping demand curve as consumers want lower, affordable prices whereas farmers would prefer higher prices.

Generally, buffer stock schemes are the majority of the time, a failure. As Extract C states ‘the schemes that failed typically involved commodities whose prices were disturbed by long-lasting shocks’. They’re generally erratic and unpredictable, therefore unreliable due to the factor that you cannot forecast a good harvest.

Evaluate the Case for and Against Buffer Stock Schemes Essay

Acid dissociation constant Essay

Acid dissociation constant Essay.

Both must be present!
A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Consider an equal molar mixture of CH3COOH and CH3COONa
CH3COOH (aq)

H+ (aq) + CH3COO- (aq)

Adding more acid creates a shift left IF enough acetate
ions are present

16.3

Which of the following are buffer systems?
(a) KF/HF
(b) KCl/HCl,
c) Na2CO3/NaHCO3

(a) KF is a weak acid and F- is its conjugate base
buffer solution
(b) HCl is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is it conjugate acid
buffer solution

16.3

Buffers
• are solutions that have the property of resisting changes in pH when acids or bases are added to them
• this property results from the presence of a buffer pair which consists of either:
 weak acid and some salt of a weak acid/its conjugate base
Ex. Acetic acid and sodium acetate  weak base and some salt of a weak base/its conjugate acid
Ex. Ammonium hydroxide and ammonium chloride

BUFFERS IN BIOLOGIC SYSTEMS
 Blood is maintained at a pH about 7.4 by the so-called primary buffers in the plasma and the secondary buffers in the
erythrocytes.
 The plasma contains carbonic acid/bicarbonate and acid/alkali sodium salts of phosphoric acid as buffers
 Plasma proteins, which behaves as acids in blood can combine with bases and so act as buffers.
 In the erythrocytes, the two buffer systems consist of
hemoglobin/oxyhemoglobin and acid/alkali potassium salts of
phosphoric acid.

Pharmaceutical Applications
(1) Preparation of such dosage forms as injections and ophthalmic solutions which are placed directly into pH-sensitive body fluids
(2) Manufacture of formulations in which the pH must be maintained at a relatively constant level to ensure maximum product stability
(3) Pharmaceutical tests and assays requiring adjustment to or maintenance of a specificp H for analytic purposes Henderson-Hasselbalch Equation
 aka: Buffer Equation
 has two forms:
 For weak acid:
pH = pKa + log [salt]
[acid]

Ex. What is the pH of a buffer solution prepared with 0.05 M sodium borate and 0.005 M boric acid?(Ka of boric acid = 5.8×10-10) pH = pKa + log [salt]
[acid]
= 9.24 + log 0.05
0.005
= 9.24 + log 10
= 9.24 + 1
= 10.24, answer

 For weak bases:
pH = pKw – pKb + log [base]
[salt]
Ex. What is the pH of a buffer solution with 0.05 M
ammonia and 0.05 M ammonium chloride? The Kb
value of ammonia is 1.80 x 10-5 at 25 o C.
Kb = 1.80 x 10-5
pKb = – log Kb
= – (1.80 x 10-5)
= – (-4.74)
pH = 14 – 4.74 + 0.05
0.05
= 9.26 + log 1
= 9.26

To calculate the Molar Ratio of Salt/Acid for a Buffer
System of Desired pH
Example:
What molar ratio of salt/acid is required to prepare a sodium acetate-acetic acid buffer solution with a pH of 5.76? The pKa value of acetic acid is 4.76 at 25o C.
pH = pKa + log [salt]
[acid]
log [salt] = pH – pKa
[acid]
= 5.76 – 4.76 = 1
Antilog of 1 = 10
Ratio = 10/1 or 10:1, answer

• Percentage of the ionized (salt) species in the
buffer system is taken as:
% ionized (salt) = ___salt__ x 100
salt + acid
Ex. The molar ratio of salt to acid is 10:1. What
is the percentage of the ionized species in the
buffer system?

% ionized (salt) = ___10__ x 100
10 + 1
= 90.91%

• Percentage of the unionized (acid) species in the
buffer system is taken as:
% unionized (acid) = ___acid__ x 100
acid + salt
Ex. The molar ratio of salt to acid is 10:1. What is
the percentage of the unionized species in the
buffer system?
% unionized (ACID) = ___1__ x 100
1 + 10
= 9.09%

Buffer Capacity
 aka:
–buffer action
– buffer efficiency
– buffer index
– buffer value
 is the ability of a buffer solution to resist changes in pH upon addition of an acid/alkali
 Approximate formula
β = ∆B
∆pH

Exact Formula/Koppel-Spiro-Van Slyke’s Equation
β = 2.3C _Ka [H3O+]__
(Ka + [H3O+])2
Maximum Buffer Capacity
 occurs when pH = pKa
βmax = 0.576 C
in which C is the total buffer concentration Ex. What is the maximum buffer capacity of an acetate buffer with a total concentration of 0.020 mole per liter?
βmax = 0.576 x 0.020
= 0.01152 or 0.012

ASSIGNMENT
1. What is the pH of a solution containing
0.30 M HCOOH and 0.52 M HCOOK?
2. What molar ratio of salt/acid
3. Percentage of the unionized (acid) species in the buffer system
4. Percentage of the ionized (SALT) species
in the buffer system

QUANTITY OF COMPONENTS IN A BUFFER SOLUTIONS TO YIELD A
SPECIFIC VOLUME

Ex. The molar ratio of sodium acetate to acetic acid

in a buffer solution with a pH of 5.76 is 10:1.
Assuming the total buffer concentration is 2.2
x10-2 mol/L, how many grams of sodium acetate
(m.w. 82) and how many grams acetic acid
(m.w.60) should be used in preparing a liter of
the solution?

SOLUTION:
m.f. sodium acetate = 10/1+10 or 10/11
m.f. Hac

= 1/1+10 or 1/11

Total buffer conc. = 2.2 x10-2 mol/L

Conc of NaAc = 10/11 x (2.2 x 10-2)
= 2 x 10-2 mol/L
Conc of HAc = 1/11 x (2.2 x 10-2)
= 0.2 x 10-2 mol/L
2 x 10-2 mol/L or 0.02 x 82 = 1.64 g of NaAc per liter of solution. 0.2 x 10-2 mol/L or 0.002 x 60 = 0.120 g of HAc per liter of solution

Change in pH with addition of Acid or base
Ex. Calculate the change in pH after adding 0.04
mol of NaOH to a liter of a buffer solution
containing 0.2 M conc. Of NaAC and Hac. The
pKa value of acetic acid is 4.76 at 250C.

pH = pka + log salt/acid
= 4.76 + log 0.2 M/0.2
= 4.76 + log of 1
= 4.76 + 0
= 4.76 (before adding NaOH)

pH = pka + log salt + base/acid – base
pH = pka + log 0.2 + 0.04/0.2 – 0.04
= pka + log 0.24/0.16
= 4.76 + 0.1761 = 4.9361 or 4.94
= 4.94 – 4.76 = 0.18 unit (after adding
NaOH)

ASSIGNMENT
1. What molar ratio of salt to acid would be required to prepare a buffer solution with a pH of 4.5? The pKa value of the acid is 4.05 at 25o C.
2. What is the change in pH on adding 0.02 mol of sodium
hydroxide to a liter of a buffer solution containing 0.5 M of sodium acetate and 0.5 M acetic acid? The pka value of acetic acid is 4.76 at 25 degrees centigrade.
3. The molar ratio of salt to acid needed to prepare a sodium acetate-acetic acid buffer solution is 1:1. Assuming that the total buffer concentration is 0.1 mol/L, how many grams of
sodium acetate should be used in preparing 2 liters of the
solution?
4. What is the change in pH with the addition of 0.01 mol HCl to a liter of buffer solution containing 0.05 M of ammonia and 0.05 M of ammonium chloride? The kb of ammonia is 1.80 x 10-5 at
25 degrees centigrade.

Acid dissociation constant Essay