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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Chemistry » Buffer Solution Problems

Buffer Solution Problems

In this article, we will solve buffer solution problems, presenting the buffer system’s purpose and composition. Practice problems and step-by-step solutions are given.

Table of Content
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A buffer is a chemical agent that helps keep the pH of a solution approximately constant, even when acids or bases are added. Buffering is vital in living systems, allowing them to maintain homeostasis or a relatively consistent internal environment. However, in this article, the main topic of discussion is buffer solution problems to help you prepare solutions without much hassle. 

What is the Purpose of a Buffer?

Buffer can be defined as a conjugate acid and conjugate base mixture that can withstand pH shifts when small amounts of strong acid and base are introduced.

  • The acid in the buffer neutralises hydroxide ions by applying a strong base.
  • The base in the buffer neutralises hydronium ions when a strong acid is applied.

Buffering Solutions with Strong Acids or Bases

Now that we have presented the purpose let’s explore what happens when we add a strong acid or base to our F-/HF buffer. Remember that the amount of F- in the solution is 0.66M x 0.1 L = 0.066 moles, while the amount of HF in the solution is 1.0 M x 0.1L = 0.10 moles. Let’s use the Henderson-Hasselbalch Approximation to double-check the pH, using moles instead of concentrations:

Log(Base/Acid) = 3.18 + log(0.066 moles F-/0.10 moles HF) = 3.00 pH = pKa + log(Base/Acid) = 3.18 + log(0.066 moles F-/0.10 moles HF) = 3.00

Let’s observe what occurs when a small amount of strong acid, such as HCl, is added. 

When we dissolve HCl in water, it separates into H3O+ and Cl-.

Because chlorine is the conjugate base of a strong acid, it is inert and does not affect pH, so we may disregard it. H3O+, on the other hand, can alter pH and react with our buffer components. The formula is as follows:

F−(aq)+H3O(aq)+⇌HF(aq)+H2O(l)

As an equivalent amount of the conjugate base reacts for every mole of H3O+ added, the reaction will continue until one or the other is practically used up. Thus, the equilibrium constant for the reaction is enormous. 

If the F- is depleted before all of the H3O+ has been consumed, the remaining H3O+ will directly alter the pH. In this case, the buffer’s capacity will be surpassed, which you want to avoid. Let’s pretend that the amount of extra H3O+ is less than the amount of existent F-; in this case, our buffer capacity will not surpass.

Problem:

When pH = pKa in a buffer, what is the base-to-acid ratio? What happens if pH = PKA + 1?

Solution:

When the base-to-acid ratio is 1, pH equals pKa because log 1 = 0. When log (base/acid) = 1, the base-to-acid ratio is 10:1.

Problem:

When enough hydrochloric acid is dissolved to make it 0.16 M HCl, determine the pH of a buffer solution of 0.5 M NH3 and 0.5 M NH4Cl? 

Solution:

Because pKa = 14 – pKb, the pKa of ammonium ions is 9.25. When 0.16 M H+ interacts with 0.15 M ammonia, 0.15 M ammonium is formed. The Henderson-Hasselbalch equation yields a pH of 8.98 when 0.65 M ammonium ion (acid) and 0.35 M residual ammonia (base) are substituted.

Problem:

To make 1.00 L of a 0.100 mol/L buffer with a pH of 5.00, how many moles of acetic acid sodium and acetate must you use?

Solution:

pH = log([A][HA]) + pKa

4.74 + log([A][HA]) = 5.00

[A][HA]) = 0.26 log([A][HA]) log([A][HA]) log([A]

[A][HA]=10.26 = 1.82 [A][HA]=10.26 = 1.82 [A][HA]=10.26

1.82 [A] = 1.82 [A] = 1.82 [A] = 1.82 [HA]

Furthermore, [A] + [HA] = 0.100 mol/L

[HA] + [HA] = 0.100 mol/L [HA] + [HA] = 1.82 [HA] + [HA] = 0.100 mol/L

0.100 mol/L [HA] = 2.82

[HA] = 0.0355 mol/L [A] = (0.100 – 0.0355) mol/L = 0.0645 mol/L

So, to make 1 L of buffer solution, combine 0.0355 mol acetic acid and 0.0645 mol sodium acetate.

What are the Components of a Buffer?

A weak conjugate acid-base pair is required to maintain a pH range efficiently. When constructing the buffer, the usage of one or the other will rely on the desired pH.

For example:

Acetic acid (a weak organic acid CH3COOH) and salt have the acetate anion (CH3COO-), such as the sodium acetate (CH3COONa).

The Pyridine (weak base with formula C5H5N) and Pyridinium Chloride salt has its conjugate acid (C5H5NH+).

Ammonia (a weak base with the formula NH3) and salt contain the ammonium cation’s conjugate acid, such as Ammonium Hydroxide (NH4OH).

Conclusion

A buffer is a mixture that can withstand pH change when acid or basic components are added. It can neutralise little amounts of added acid or base, allowing the solution’s pH to remain relatively constant. Buffer solutions have a functional pH range and a capacity to determine how much acid/base can be neutralised before the change in the pH.

faq

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

Why is it so important to keep our blood pH balanced?

Ans. Believe it or not, if the pH of our blood falls below 6.8 or rises beyond 7.8, body cells can cease working, an...Read full

What causes buffers to lose their efficacy?

Ans. The balanced equation and the formula for the equilibrium acid dissociation constant can be used to calculate t...Read full

Explain why a buffer’s pKa should be as close to the desired pH as possible?

Ans. The pKa should be close to the target pH such that the Henderson-Hasselbalch equation’s base-to-acid ratio is...Read full

Ans. Believe it or not, if the pH of our blood falls below 6.8 or rises beyond 7.8, body cells can cease working, and the individual can die. This demonstrates the significance of maintaining a healthy blood pH. Enzymes are quite specific and work best at the appropriate temperature and pH. If the blood pH falls outside of this range, the enzymes cease operating. And in certain cases, enzymes can become irreversibly denatured, preventing them from catalysing reactions. This, in turn, impacts several basic processes in the human body, resulting in various disorders.

Ans. The balanced equation and the formula for the equilibrium acid dissociation constant can be used to calculate the change in pH of a buffer when an acid or base is added. If too much strong acid or base is applied, any buffer will lose its effectiveness.

Ans. The pKa should be close to the target pH such that the Henderson-Hasselbalch equation’s base-to-acid ratio is near 1. As the base-to-acid ratio deviates from 1, adding acids and bases to the buffer will significantly impact the pH.

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