A buffer solution is a mixture of a weak acid and its conjugate base. It resists any change in the pH upon the addition of acidic or basic components. We can classify buffer solutions into two kinds: Acidic and basic buffer solutions.
An acidic buffer solution is a mixture of a weak acid and its conjugate strong base. In contrast, a basic buffer solution is a mixture of a weak acid and its conjugate base.
| Definition: A buffer solution is a mixture of a weak acid and its conjugate base. It resists any change in the pH upon the addition of acidic or basic components. |
Also Read: Buffer Chemistry
Buffer Chemistry Questions with Solutions
Q1. Why does a buffer solution resist any change in pH?
(a) They give unionised acid or base on reaction with acid or a base.
(b) Acids and bases in the buffer solutions are protected from attack by other ions
(c) They have an excess of H+ or OH– ions.
(d) They have a fixed pH value.
Answer: (a) A buffer solution resists any change in pH because they give unionised acid or base on reaction with acid or a base.
Q2. Which of the following mixture is an example of a buffer solution?
(a) NaNO2 and HNO2
(b) KCl and HCl
(c) NH4NO3 and HNO3
(d) NaCl and NaOH
Answer: (a) NaNO2 and HNO2 mixture is an example of the buffer solution.
Q3. Which of the following mixture is not an example of an acidic buffer solution?
(a) Na2CO3 and H2CO3
(b) CH3COONa and CH3COOH
(c) NaClO4 and HClO4
(d) Na3PO4 and H3PO4
Answer: Mixture of NaClO4 and HClO4 is not an example of an acidic buffer solution.
Q4. A buffer solution is a mixture of
(a) Weak Acid and Strong Base
(b) Strong Acid and Weak Base
(c ) Strong Acid and its conjugate base
(d) Weak Acid and its conjugate base
Answer: (d) Buffer solution is a mixture of a weak acid and its conjugate base.
Q5. What is the pH of a buffer solution?
(a) Same as its pKa value
(b) Same as its Ka value
(c ) Can’t be calculated
(d) None of the above
Answer: (b) The pH of a buffer solution is the same as its pKa value.
Q6. What is a buffer solution? Give an example of a buffer solution.
Answer: A buffer solution is a mixture of a weak acid and its conjugate base. It resists any change in the pH upon the addition of acidic or basic components.
Example: A mixture of acetic acid and sodium acetate.
Q7. What are the various types of buffer solutions?
Answer: There are two types of buffer solutions.
1. Acidic buffer solution
2. Basic Buffer solution
Q8. What is a basic buffer solution? Give an example of a basic buffer solution.
Answer: A basic buffer solution is a mixture of a weak base and its conjugate strong acid. It resists any change in the pH upon the addition of acidic or basic components.
Example: A mixture of ammonium hydroxide and ammonium chloride.
Q9. What is Henderson and Hasselbalch Equation? Give one limitation of the Handerson Equation.
Answer: Henderson and Hasselbalch proposed an equation that gives a relation between the pH or pOH and pKa or pKb and the ratio of the concentrations of the ionised chemical species.
pH = pKa + log10 ([A–] / [HA])
Here,
[A–] refers to the molar concentration of the conjugate base of the acid [HA] refers to the molar concentration of the weak acid.Limitation of Henderson and Hasselbalch Equation: It cannot be used for strong acids and strong bases.
Q10. Distinguish between an acidic buffer solution and a basic buffer solution.
Answer:
| S. No. | Acidic Buffer Solution | Basic Buffer Solution |
| 1. | An acidic buffer solution is a mixture of a weak acid and its conjugate strong base. | A basic buffer solution is a mixture of a weak base and its conjugate strong acid. |
| 2. | Example: A mixture of acetic acid and sodium acetate. | Example: A mixture of ammonium hydroxide and ammonium chloride. |
| 3. | Its pH is less than 7. | Its pH is more than 7. |
Q11. Match the following
| Column A | Column B |
| Acid | A substance with one more proton than a base. |
| Base | A mixture of a weak acid or base and its salt. |
| Conjugate Acid | A substance with one less proton than an acid. |
| Conjugate Base | A substance that acts as a proton acceptor. |
| Buffer Solution | A substance that acts as a proton donor. |
Answer:
| Column A | Column B |
| Acid | A substance that acts as a proton donor. |
| Base | A substance that acts as a proton acceptor. |
| Conjugate Acid | A substance with one less proton than an acid. |
| Conjugate Base | A substance with one more proton than a base. |
| Buffer Solution | A mixture of a weak acid or base and its salt. |
Q12. What is buffering capacity?
Answer: Buffering capacity is the number of millimoles of an acid or a base to be added to one litre of buffer solution to change its pH by a unit.
Βuffering capacity = Millimoles / (Δ pH)
Q13. What is the ratio of base and acid when the pH of the solution is equivalent to the pKa in the buffer solution? How will the result alter if the pKa is increased by unity?
Answer: According to the Henderson and Hasselbalch equation:
pH = pKa + log10 ([A–] / [HA])
Given that the pH of the solution is equal to the pKa.
It is only possible when the ratio of base to acid is equivalent to 1 as log 1 = 0.
If pKa is increased by unity, then the log (base/acid) must be equivalent to 1. Thus, the ratio of base to acid must be 10:1.
Q14. 100 ml of 0.1 M CH3​COOH is mixed with 50 ml of 0.1 M NaOH solution, and the pH of the resulting solution is 5. What is the change in pH if 100 ml of 0.05 M NaOH is added to the above solution?
Answer: According to the Henderson and Hasselbalch equation:
pH = pKa + log10 ([A–] / [HA])
pH = pKa + log10 (0.1 X 0.05​ / 0.1 X 0.05)
5 = pKa
When 100 ml of 0.05 M NaOH is added, the acid is completely neutralized, to form sodium acetate salt.
The expression for the hydrogen ion concentration of a weak acid and strong base salt.
pH = ½ (pKw​ + pKa ​+ log c)
pH = ½ (14 + 5 ​+ log (0.1 / 0.25)
pH = 8.8.
Hence, the change in pH will be
Δ pH = 8.8 − 5
Δ pH = 3.8.
Q15. A specific buffer solution contains an equal concentration of X− and HX. The Kb​ for X− is 10−10. What is the pH of the buffer solution?
Answer: According to the Henderson and Hasselbalch equation:
pH = pKa + log10 ([A–] / [HA])
pOH = pKb + log10 ([A–] / [HA])
Here the concentration of [A–] is equivalent to the [HA]
log10 ([A–] / [HA]) = log10 1
log10 ([A–] / [HA]) = 0
Putting value in the Henderson and Hasselbalch equation.
pOH = pKb
pOH = log10−10
pOH = 10
Further, pH + pOH = 14.
Putting the value of pOH in the above equation.
pH + 10 = 14
pH = 14 – 10
pH = 4.
Practise Questions on Buffer
Q1. A 0.1 mole of CH3​NH2​ with Kb​ 5 X 10−4 is mixed with 0.08 mole of HCl and diluted to one litre. What is the [H+] concentration in the solution?
Q2. The pH of 0.1M solution of the following salts increases in the order of:
(a) NaCl < NH4​Cl < NaCN < HCl
(b) HCl < NH4​Cl < NaCl < NaCN
(c ) NaCN < NH4​Cl < NaCl < HCl
(d) HCl < NaCl < NaCN < NH4​Cl
Q3. What is [H+] in mol / L of a solution 0.20 M in CH3​COONa and 0.10 M in CH3​COOH? The Ka​ for CH3​COOH is 1.8 X 10−5.
Q4. In a buffer solution containing an equal B− and HB concentration, the Kb​ for B− is 10−10. What is the pH of the buffer solution?
Q5. The pH of a buffer solution containing 0.1 M CH3​COOH and 0.1 M CH3​COONa is 4.74. If 0.05 mole of HCl is added to one litre of this buffer solution, what will be the pH? Given Ka of CH3​COOH is equal to 1.8×10−5.
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