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>> Buffer Capacity-Part1
The property of buffer solution to resist alteration in its pH value is known as buffer capacity. It has been found that if the ratio [Salt]/[Acid] or [Salt]/[Base] is unity, the pH of a particular buffer does not change at all. Buffer capacity is defined quantitatively as number of moles of acid or base added in one litre of solution as to change the pH by unity,
(φ) = (No.of moles of acid or base added to 1 litre)/(Change in pH)
or φ = δb/δ(ph)
where δb → number of moles of acid or base added to 1 litre solution and δ(pH) → change in pH.
Buffer capacity is maximum:
(i) When [Salt] = [Acid], i.e., pH = pKa for acid buffer
(ii) When [Salt] = [Base], i.e., pOH = pKb for base buffer under above conditions, the buffer is called efficient.
Utility of buffer solutions in analytical chemistry
Buffers are used:
(i) To determine the pH with the help of indicators.
(ii) For the removal of phosphate ion in the qualitative inorganic analysis after second group using CH3COOH + CH3COONa buffer.
(iii) For the precipitation of lead chromate quantitatively in gravimetric analysis, the buffer, CH3COOH + CH3COONa, is used.
(iv) For precipitation of hydroxides of third group of qualitative analysis, a buffer, NH4Cl + NH4OH, is used.
(v) A buffer solution of NH4Cl, NH4OH, and (NH4)2CO3 is used for precipitation of carbonates of fifth group in qualitative inorganic analysis.
(vi) The pH of intracellular fluid, blood is naturally maintained. This maintenance of pH is essential to sustain life because, enzyme catalysis is pH sensitive process. The normal pH of blood plasma is 7.4. Following two buffers in the blood help to maintain pH (7.4).
(a) Buffer of carbonic acid (H2CO3 and NaHCO3)
(b) Buffer of phosphoric acid (H2P04, HPO2-)
Buffers are used in industrial processes such as manufacture of paper, dyes, inks, paints, drugs, etc. Buffers are also employed in agriculture, dairy products and preservation of various types of foods and fruits.
Example 40: The pH of a buffer is 4.745. When 0.01 mole of NaOH is added to 1 litre of it, the pH changes to 4.832. Calculate its buffer capacity.
Solution: From definition,
Buffer capacity (φ ) = δb/δ[ph]
δ (pH) = (4.832 - 4.745) = 0.087; δb = 0.01
Substituting given values,
φ = 0.01/0.087 = 0.115
Example 41: Suppose it is required to make a buffer solution of pH = 4, using acetic acid and sodium acetate. How much of sodium acetate is to be added to 1 litre of N/10 acetic acid?
Solution: Applying Henderson's equation,
pH = log - log Ka
4 = log [Salt] - log (0.1) - log 1.8 × 10-5
So log [Salt] = (4 - 1 - 5 + 0.2552) = 2.2552
The molecular mass of CH3COONa = 82
Amount of salt = 0.018 × 82 = 1.476 g
Example 42: What is the pH of the solution when 0.2 mole of hydrochloric acid is added to one litre of a solution Containing 1 M acetic acid and acetate ion? Assume that the total volume is one litre. Ka for CH3COOH = 1.8 × 10-5.
Solution: On adding HCl, the free hydrogen ions will combine with CH3COO- ions to form CH3COOH. Thus, the concentration of acetic acid increases while that of CH3COO- ions decreases.
[CH3COOH] = (0.2 + 1) = 1.2 mol litre-1
[Salt] = (1 - 0.2) = 0.8 mol litre-1
Applying Henderson's equation,
pH = log [Salt]/[Acid] - log Ka
= log0.8/1.2 log 1.8 ×10-5
= log 2 - log 3 - log 1.8 × 10-5 = 4.5687
Example 43: 20 mL of 0.2 M sodium hydroxide is added to 50 mL of 0.2 M acetic acid to give 70 mL of the solution. What is the pH of the solution? Calculate the additional volume of 0.2 M NaOH required to make the pH of solution 4.74. The ionisation constant of acetic acid is 1.8×10-5.
Solution: No. of moles of NaOH in
20 mL = 0.2/1000 = 0.004 02
No. of moles of acetic acid in 50 mL = 0.2/1000 x 50 = 0.01
When NaOH is added, CH3COONa is formed
CH3COOH + NaOH ↔ CH3COONa + H2O
1 mole 1 mole mole 1 mole
No. of moles of CH3COONa in 70 mL solution = 0.004
No. of moles of CH3COOH in 70 mL solution
= (0.01 - 0.004) = 0.006
Applying Henderson's equation,
pH = log[Salt]/[Acid] log Ka
= log(0.004/0.006) log 1.8 × 10-5 = 4.56787
On further addition of NaOH, the pH becomes 4.74.
pH = log[Salt]/[Acid] - log Ka
= log[Salt]/[Acid] - log 1.8 × 10-5
or log[Salt]/[Acid] = pH+log1.8×10-5=(4.74-4.7448)= -0.0048
So log[Salt]/[Acid] = 1.9952
[Salt]/[Acid] = -.9891
Let 'x' moles of NaOH be added.
[Salt] = (0.004 + x) mole
[Acid] = (0.006 - x) mole
[Salt]/[Acid] = (0.004+x)/(0.006-x) = 0.9891
or 0.004 + x = 0.9891 × 0.006 - 0.9891x
x = 0.000972 mole
Volume of 0.2 M NaOH solution having 0.000972 mole
= 1000/0.2×.000972=4.86 mL
Example 44: Calculate pH of the buffer solution containing 0.15 mole of NH4OH and 0.25 moles of NH4Cl. Kb for NH4OH is 1.98 × 10-5.
Solution: Applying the equation,
pOH = log[Salt]/[Base] - log Kb
= log(0.25/0.15) - log 1.8 × 10-5
= klog 5 - log 3 - log 1.8 × 10-5
= 0.6989 - 0.4771 + 4.7448 = 4.966
pH = (14 - 4.966) = 9.034
Example 45 : What volume of 0.10 M sodium formate solution should be added to 50 mL of 0.05 M formic acid to produce a buffer solution of pH 4.0? pKa for formic acid is 3.80.
Solution: Let x mL of 0.10 M sodium formate be added.
No. of moles in x mL of 0.10 M sodium formate = 0.10/1000 ×x
No. of moles in 50 mL of 0.05 M formic acid = 0.05/1000 * 50
[Sod.formate]/[formicacid]=((1.10×x)/1000)/((0.05×50)/1000) (0.10 x)/2.5 = 0.04 x
Applying the equation
pH = log[NaCN]/[HCN] - log Ka
8.5 = log (0.01-a)/a - log4.1 *10-10
So, log(0.01-a)/a = 8.5 + 0.6127 - 10.0 = 1.1127
(0.01-a)/a = 0.1296
or a = 0.01/1.1296 = 0.0089 mole