Question:medium

20 mL of 0.1 M NH4OH is mixed with 40 mL of 0.05 M HCl.The pH of the mixture is nearest to:(Given: Kb(NH4OH) = 1 × 10–5 ,log 2 = 0.30,log 3 = 0.48, log 5 = 0.69, log 7 = 0.84, log 11 =1.04)

Updated On: Mar 20, 2026
  • 3.2

  • 4.2

  • 5.2

  • 6.2

Show Solution

The Correct Option is C

Solution and Explanation

To find the pH of the mixture formed by mixing 20 mL of 0.1 M NH4OH with 40 mL of 0.05 M HCl, we need to follow these steps:

  1. Calculate the millimoles of NH4OH and HCl:
    • Millimoles of NH4OH = Volume (mL) × Molarity (M) = 20 × 0.1 = 2 mmol
    • Millimoles of HCl = Volume (mL) × Molarity (M) = 40 × 0.05 = 2 mmol
  2. Determine the reaction between NH4OH and HCl:
    • The reaction: NH4OH + HCl → NH4Cl + H2O
    • This is a neutralization reaction, where 1 mole of NH4OH completely reacts with 1 mole of HCl to form NH4Cl.
    • Since initial millimoles of NH4OH and HCl are equal, they completely neutralize each other, forming 2 mmol of NH4Cl.
  3. Since a weak base (NH4OH) reacts with a strong acid (HCl), the resulting solution contains the salt NH4Cl, which slightly hydrolyzes to affect pH.
  4. To calculate the pH of the resultant weak acid (from NH4Cl hydrolysis):
    • The hydrolysis of NH4Cl in water is primarily influenced by the Ka of its conjugate acid NH4+:
    • Ka(NH4+) = \frac{K_w}{K_b} where K_w = 1 \times 10^{-14}
    • Using given Kb(NH4OH) = 1 × 10–5, Ka(NH4+) = \frac{1 \times 10^{-14}}{1 \times 10^{-5}} = 1 \times 10^{-9}
  5. Apply the formula to calculate pH:
    • Total Volume = 20 mL + 40 mL = 60 mL
    • [NH4+] = \frac{2 \text{ mmol}}{60 \text{ mL}} = \frac{1}{30} \text{ M}
    • Using pH formula for weak acid salt: pH = 7 + 0.5 log \frac{K_w}{[NH_4^+] \cdot K_a}
    • Substitute the values: 7 + 0.5 \cdot \log \left(\frac{1 \times 10^{-14}}{\left(\frac{1}{30}\right) \cdot 1 \times 10^{-9}}\right)
    • This simplifies to: 7 + 0.5 \cdot \log(30 \times 10^{-5})
    • Calculate: \log 30 \approx \log(3 \times 10) = \log 3 + \log 10 = 0.48 + 1 = 1.48
    • pH = 7 + 0.5 × (1.48 - 5) = 7 + 0.5 × -3.52 = 7 - 1.76 = 5.24

Therefore, the pH of the mixture is nearest to 5.2.

Was this answer helpful?
7

Top Questions on Equilibrium


Questions Asked in JEE Main exam