Buffer Solutions
At ordinary temperature, the pH of pure water is 7. Now, when 1 mL of (M) HCl solution is added to 100 mL of pure water, it is observed that the value of pH decreases from 7 to 2.
Again, if 1 mL of 1 (M) NaOH solution is added to 100 mL of pure water the value of the pH of the solution increases from 7 to 2. However, many solutions resist the change in pH even when a small quantity of acid or base is added to them. Such solutions are called buffer solutions.
Buffer Solutions Definition:
A buffer solution may be defined as a solution that resists the change in its pH when a small amount of acid or base is added to it.
Various types of buffer solutions depending on the nature of the acid/base and the salt:
1. Solution of a weak acid and its salt
An aqueous solution of a weak acid and its salt acts as a buffer solution.
Example:
Aqueous solutions of CH3COOH and CH3COONa, H2CO3 and NaHCO3, citric acid and sodium citrate, boric acid, and sodium borate, etc.
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2. Solution of a weak base and its salt
An aqueous solution of a weak base and its salt acts as a buffer solution.
Example: Aqueous solutions of NH3 and NH4Cl, aniline and anilinium hydrochloride, etc.
3. Solution of two salts of a polybasic acid
An aqueous solution containing two salts of polybasic acid can act as a buffer solution.
Example:
Aqueous solutions of Na2CO3 and NaHCO3 (NaHCO3 is a weak acid and Na2CO3 is its salt), NaIT2PO4 and Na2HPO4 etc.
4. Solution of a salt derived from a weak acid and a weak base
An aqueous solution of a salt formed by a weak acid and a weak base can function as a buffer solution.
Example:
CH3COONH4 is a salt of weak acid (CH3COOH) and weak base (NH4OH). A solution of this salt acts as a buffer solution.
Types of buffer solutions depending on their pH values: Depending on pH values, buffer solutions are of two types—
1. Acidic buffer
Buffer solutions with a pH lower than 7 are called acidic buffers. Aqueous solutions of CH3COOH and CH3COONa, aqueous solutions of lactic acid and sodium lactate, etc., are some common examples of acidic buffer solutions.
2. Basic buffer
Buffer solutions with a pH higher than 7 are called basic buffers. An aqueous solution of NH4OH and NH4Cl is an example of a basic buffer solution.
Mechanism of buffer action
Mechanism of buffer action Definition:
The ability of a buffer solution to resist the change of its pH value on the addition of a small amount of an acid or a base to it is called buffer action.
Mechanism of action of an acidic buffer
To understand the mechanism of buffer action of an acidic buffer, let us consider an acidic buffer solution that consists of CH3COOH and CH3COONa.
In the solution, CH3COONa almost completely dissociates into CH3COO–(aq) and Na+ ions, whereas acetic acid, being a weak electrolyte, partially dissociates into CH3COO-(aq) and H3O+(aq).
The partial dissociation of acetic acid leads to the formation of the following equilibrium.
CH3COOH(aq) + H2O (l) ⇌ CH3COC–(aq) + H3O+(aq)
As CH3COOH ionizes partially and CH3COONa ionizes almost completely, the solution consists of high concentrations of CH3COOH and CH3COO–ions
Addition of a small amount of strong acid:
If a small amount of strong acid (example: HCl ) is added to this buffer solution, H3O+ ions produced from the strong acid combine with an equal number of CH3COO- ions present in the solution to form unionized CH3COOH molecules:
CH3COO–(aq)+ H3O+(l)→CH3COOH(aq) + H2O(l)
This causes the equilibrium involved in the ionization of CH3COOH (equation 1) to shift to the left. As a result, the concentration of H3O+ Ionsin in the solution virtually remains the same, i.e., the pH of the solution remains almost unchanged.
Addition of a small amount of strong base:
When a small amount of a strong base like NaOH, OH, etc., is added to this buffer solution, the OH– ions obtained from the strong base combine with an equal number of H4, and ions (which results mainly from the dissociation of CH3COOH ) present in the solution, producing unionized water molecules.
Consequently, the equilibrium associated with the ionization of CH3COOH [eqn. (1)] gets disturbed. So, according to Le Chatcller’s principle, some molecules of CH3COOH ionize to produce H3O– ions, and the equilibrium, shifts towards the right. Therefore, the addition of a small quantity of a strong base with almost no change in are to the solution causes the concentration of either OH– ions or H2O+ ions, and consequently pH of the solution remains almost unaltered.
Mechanism of action of basic buffer:
Let us consider a basic buffer solution consisting of a weak base NH3 and its salt, NH4Cl. Being a strong electrolyte, ammonium chloride (NH4Cl) dissociates almost completely in solution:
NH4Cl(aq)→ NH4–(aq) + Cl–(aq).
However, NH3, being a weak base, reacts partially with water and forms the following equilibrium solution.
⇒ \(\mathrm{NH}_3(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_4^{+}(a q)+\mathrm{OH}^{-}(a q)\) …………………………………………..(1)
As NH3 reacts partially with water and NH4Cl gets completely dissociated, the concentration of NH3 and NH3 in the solution is sufficient.
Addition of a small amount of strong acid:
If a small quantity of strong acid (for example HCl, H2SO4) is added to this buffer solution, H3O+ ions obtained from the strong acid react almost completely with an equal number of OH- ions (which results mainly from the reactions of NH3 with water) present in the solution and produce unionized water molecules.
As a result, the equilibrium [eqn. (1)] gets disturbed. According to Le Chatelier’s principle, some NH3 molecules react with water to form OH– ions. As a result, the equilibrium [eqn. (1)] shifts to the right. So, the addition of a small amount of a strong acid to the buffer solution causes almost no change in concentration of either OH– ions or H3O+ ions, i.e., the pH of the solution remains unchanged.
Addition of a small amount of strong base:
When a small quantity of a strong base (for example NaOH, KOH is added to the buffer solution, OH- ions produced by the strong base react almost completely with an equal number of NHJ ions present in the solution and form unionized
NH3 molecules [NH+(aq) + OH– (aq) →NH3 (aq) + H2O(l)].
As a result the equilibrium [eqn. (1)] shifts to the left. Consequently, the concentration of OH- ions in the solution virtually remains the same, and hence pH of dissolution remains unchanged.
Buffer action of two salts of polybasic acid (in solution):
Let us consider the buffer solution comprising of two salts NaH2PO4 and Na2HPO4 (which are the salts of polybasic acid, H3PO4 ). In this solution, NaH2PO4 acts as an acid while Na2HPO4 as a salt, and both of them undergo complete dissociation as given below:
NaH2PO4(aq) → Na+(aq) + H2PO4– (aq)
Na2HPO4(aq)→ 2Na+(aq) + HPO42-(aq)
H2PO4 itself is a weak acid and due to a common ion (HPO42-), it undergoes slight ionization in the solution forming the following equilibrium:
⇒ \(\mathrm{H}_2 \mathrm{PO}_4^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{HPO}_4^{2-}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q)\) ……………………(1)
As H2PO4 undergoes slight ionization and Na2HPO4 gets completely ionized, the concentration of H2PO4 and HPO4– in the solution is sufficient.
Addition of a small amount of strong acid:
When a small quantity of a strong acid is added to this buffer solution, H3O+ ions produced by the strong acid react almost completely with an equal number of HPO– ions, and form unionized H2PO4 ions
⇒ \(\mathrm{HPO}_4^{2-}(a q)+\mathrm{H}_3 \mathrm{O}^{+}(a q) \rightarrow \mathrm{H}_2 \mathrm{PO}_4^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l)\)
As a result, the equilibrium [eqn. (1)] shifts towards the left. Hence, the effect of the addition of a small amount of strong acid is neutralized. Thus, the pH of the solution remains unchanged.
Addition of a small amount of strong base:
When a small quantity of a strong base is added to the buffer solution, OH– ions from the added base almost completely react with an equal number of H3O+ ions (which results mainly from the ionization of H2PO4 ) present in the solution and form unionized water molecules, This disturbs the ionization equilibrium of NaH2PO4.
As a result, according to Le Chatelier’s principle, some molecules of NaH2PO4 ionize to form H3O+ ions and cause the equilibrium to shift towards the right. Therefore, the addition of a small quantity of a strong base makes no change in the concentration of either OH– ions or H3O+ ions, and consequently, the pH of the solution remains almost unchanged.
Determination of pH of a buffer solution: Henderson’s equation
Let us consider a buffer solution consisting of a weak acid (HA) and its salt (MA). In the solution, MA dissociates completely, forming M+(aq) and A–(aq) ions, while the weak acid, HA, because of its partial ionization, forms the following equilibrium
⇒ \(\mathrm{HA}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(a q)+\mathrm{A}+(a q)\)
Applying the law of mass action to the abbveÿequilibrium, we have the ionization constant of HA
⇒ \(K_a=\frac{\left[\mathrm{H}_3 \mathrm{O}^{+}\right] \times\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]},\)
Where, [H3O+], [A–], and [HA] are the molar concentrations of H3O+, A–, and HA respectively at equilibrium.
∴ \(\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=K_a \times \frac{[\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}\)
Taking negative logarithms on both sides we get,
⇒ \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log _{10} K_a-\log _{10} \frac{[\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}\)
⇒ \(-\log _{10}\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=-\log _{10} K_a-\log _{10} \frac{[\mathrm{HA}]}{\left[\mathrm{A}^{-}\right]}\)
Or, ,\(p H=p K_a+\log _{10} \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)………………(1)
As HA is a weak acid, its ionization in the solution is very small, which gets further decreased in the presence ofthe common ion, A–. Therefore, the molar concentration of unionized HA can be considered to be the same as the initial molar concentration of HA.
Again, the molar concentration of A– ions produced by complete dissociation of MA is much higher than that of A– ions produced by partial ionization of HA. Therefore, the total molar concentration of A– ions in the solution is almost the same as the initial molar concentration of MA in the solution.
∴ From equation (1) \(p H=p K_a+\log \frac{[\text { salt }]}{[\text { acid }]}\) ………………(2)
Where Ka is the ionization constant of the weak acid; [acid] and [salt] are the initial molar concentrations of acid and salt respectively the buffer solution. Equation [2] is called the Henderson’s equation.
It is used to determine the pH of a buffer solution consisting of a weak acid and its salt. Similarly, the equation for determining the pOH of a buffer solution consisting of a weak base and its salt is-
⇒ \(p O H=p K_b+\log \frac{[\text { salt] }}{\text { [base] }}\)
Where, K, b is the ionization constant ofthe weak base; [salt] and [base] represent the initial molar concentrations of salt and base respectively. The above equation can be rewritten as,
⇒ \(p H=14-p O H=14-p K_b-\log \frac{[\text { salt] }}{\text { [base] }}\)
At a fixed temperature, the value of pKa of a weak acid is fixed. Hence, at a fixed temperature, the pH of a buffer solution consisting of a weak add and its salt depends upon the die value of the pKa of the weak acid as well as the ratio of the molar concentrations of the salt to the molar concentration of the acid.
Similarly, at a particular temperature, the pH of a buffer solution consisting of a weak base and its salt depends on the value of pKb of the weak base and the ratio of molar concentrations of the salt to the base.
Applications of Henderson’s equation:
If the molar concentrations of a weak acid (or base) and its salt present in a buffer solution as well as the dissociation constant of that acid (or base) are known, then that solution can be determined by using Henderson’s equation.
If in an acidic or basic buffer solution, the molar concentrations of the weak acid (or base) and its salt and pH of that solution are known, then the dissociation constant of the weak acid (or base) can be calculated with the help ofHenderson’s equation.
For the preparation of a buffer solution with a desired value of pH, the ratio of the weak acid and its salt (or weak base and its salt) in the solution can be determined by Henderson’s equation.
Buffer Capacity
Buffer Capacity Definition:
The buffer capacity of a buffer solution is defined as the number of moles of a strong base or an acid required to change the pH of 1L of that buffer solution by unity.
When an acid is added to a buffer solution, its pH value decreases, whereas when a base is added to a buffer solution, its pH value increases.
Mathematical explanation:
When the pH of 1 L of solution increases by d(pH) due to the addition of db mol of any strong base, then the buffer capacity of that buffer solution
⇒ \(\beta=\frac{d b}{d(p H)}\)
Similarly, when the pH of 1L of buffer solution decreases by d(pH) due to the addition of da mol of any acid, then the buffer capacity of that buffer solution \(\beta=\frac{d a}{d(p H)}\)
Some important facts regarding buffer capacity:
- If the buffer capacity of any buffer solution is high, then a greater amount of a strong add or base will be required to bring about any change in the pH of that solution.
- Between two buffer solutions having the same components, the buffer capacity of that solution will be higher when the concentrations of the components are
higher. - In a buffer solution, if the difference in molar concentration of die components is small, then the addition of a small amount of strong acid or base causes a small change in molar concentrations of die components.
Consequently, the change in pH of the die solution also becomes small. For this reason, the die buffer capacity of a buffer solution becomes maximum when the molar concentrations ofthe components are the same.
1. The buffer capacity of a buffer solution consisting of a weak acid and its salt becomes maximum when die molar concentrations of the weak acid and its salt become equal. Under this condition, the pH of the buffer solution = pKa of the weak acid
Example: The buffer capacity of CH3COOH/CH3COONa solution will be maximum when its pH = 4.74 (since, pKa of CH3COOH =4.74).
2. For the same reason, the buffer capacity of a buffer solution consisting of a weak base and its salt becomes maximum when the pOH of the buffer becomes equal to the pKb of the weak base. For example, the buffer capacity of NH3/NH4Cl solution will be maximum when its pOH = 4.74 (since, pKb of NH3 = 4.74)
pH range of buffer capacity:
A buffer solution consisting of a weak acid and its salt will work properly only when the ratio of the molar concentration of the salt to the acid lies between 0.1 to 10. Therefore, according to Henderson’s equation, the pH -range for the buffer capacity of such a buffer will vary from (pKa – 1) to (pKb + 1). 0 Similarly the range of pOH for buffer capacity of a buffer consisting of a weak base and its salt will vary from (pKb – 1) to (pKb+ 1).
Importance of butter solutions
1. The pH of human blood lies between 7.35 to 7.45, i.e., human blood is slightly alkaline. In spite, ofthe presence of acidic or basic substances produced due to various metabolic processes, the pH of human blood remains almost constant because ofthe presence of different buffer systems like bicarbonate-carbonic acid buffer (HCO–3 / H2CO3), phosphate buffer (HPO42- /H4PO4–), etc.
Excess acid in the blood is neutralized by the reaction:
HCO3– + H3O+⇌ H2CO3 +H2O. H2CO3 formed decomposes into CO2 and water.
The produced CO2 is exhaled out through the lungs. Again, if any base (OH–) from an external source enters the blood, it gets neutralized by the reaction:
H2CO3 + OH– ⇌ HCO3– +H2O.
2. Buffers find extensive use in analytical works as well as in chemical industries. In these cases, a buffer solution is used to maintain pH at a certain value. For instance, in qualitative analysis, in electroplating, tanning of hides and skins, fermentation, manufacture of paper, ink, paints, and dyes, etc., pH is strictly maintained. Buffer solution is also used for the preservation of fruits & products derived from fruits.