AIIMS 2006 Chemistry Buffer Solutions MCQ Question

To determine the pH of the mixture of ammonia and hydrochloric acid (HCl), we first need to analyze the reaction between these two substances.

Step 1: Identify the Reaction

Ammonia ($\text{NH}_3$) is a weak base, and hydrochloric acid (HCl) is a strong acid. When they are mixed, they undergo a neutralization reaction:

$$\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4^+ + \text{Cl}^-$$

Step 2: Calculate Moles of Reactants

First, we need to calculate the moles of ammonia and HCl in the respective solutions.

• Ammonia:

  $$\text{Volume of NH}_3 = 40 \, \text{ml} = 0.040 \, \text{L}$$ $$\text{Concentration of NH}_3 = 0.1 \, \text{M}$$ $$\text{Moles of NH}_3 = 0.1 \times 0.040 = 0.004 \, \text{mol}$$

• Hydrochloric acid:

  $$\text{Volume of HCl} = 20 \, \text{ml} = 0.020 \, \text{L}$$ $$\text{Concentration of HCl} = 0.1 \, \text{M}$$ $$\text{Moles of HCl} = 0.1 \times 0.020 = 0.002 \, \text{mol}$$

Step 3: Determine Limiting Reactant

In the reaction, HCl will react with NH₃. We have:

• Moles of NH₃ = 0.004 mol
• Moles of HCl = 0.002 mol

Since HCl is the limiting reactant, it will completely react with an equivalent amount of NH₃. Thus, the moles of NH₃ that remain after the reaction will be:

$$\text{Remaining moles of NH}_3 = 0.004 - 0.002 = 0.002 \, \text{mol}$$

Step 4: Calculate Concentrations in the Mixture

The total volume of the resulting solution after mixing is:

$$\text{Total Volume} = 40 \, \text{ml} + 20 \, \text{ml} = 60 \, \text{ml} = 0.060 \, \text{L}$$

Now, we can calculate the concentrations of the remaining NH₃ and the produced NH₄⁺:

• Concentration of NH₃:

$$[\text{NH}_3] = \frac{0.002 \, \text{mol}}{0.060 \, \text{L}} = 0.0333 \, \text{M}$$

• Concentration of NH₄⁺: Since HCl reacted completely, the moles of NH₄⁺ produced will be equal to the moles of HCl that reacted:

$$[\text{NH}_4^+] = \frac{0.002 \, \text{mol}}{0.060 \, \text{L}} = 0.0333 \, \text{M}$$

Step 5: Use the Henderson-Hasselbalch Equation

The solution now contains a weak base (NH₃) and its conjugate acid (NH₄⁺). We can use the Henderson-Hasselbalch equation to find the pH of the buffer solution:

$$\text{pH} = \text{pK}_a + \log\left(\frac{[\text{base}]}{[\text{acid}]}\right)$$

Given that the pKₐ of ammonia is 4.74, we have:

$$\text{pH} = 4.74 + \log\left(\frac{0.0333}{0.0333}\right)$$

Since the concentrations of NH₃ and NH₄⁺ are equal, the logarithmic term becomes:

$$\log(1) = 0$$

Thus:

$$\text{pH} = 4.74 + 0 = 4.74$$

Conclusion

The pH of the mixture is $4.74$, which corresponds to option A.

Clarification of Other Options

• **Option B (2