Verified Solution • 2025–26 Session moderate

QJEEM24Ch9363: JEE Main 2024 Chemistry - Class 11 JEEM24Ch9363 Solution: for is and for is . The of ammonium acetate solution will be…

Ka\mathrm{K}_{\mathrm{a}} for CH3COOH\mathrm{CH}_3 \mathrm{COOH} is 1.8×10−51.8 \times 10^{-5} and Kb\mathrm{K}_{\mathrm{b}} for NH4OH\mathrm{NH}_4 \mathrm{OH} is 1.8×10−51.8 \times 10^{-5} . The pH\mathrm{pH} of ammonium acetate solution will be _________.

Step-by-Step Verified Solution

To find the pH of an ammonium acetate solution, we use the fact that ammonium acetate is a salt resulting from the neutralization of a weak acid (acetic acid, CH3COOHCH_3COOH ) by a weak base (ammonium hydroxide, NH4OHNH_4OH ). The respective ionization constant values for the weak acid KaK_a and the weak base KbK_b are given as 1.8×10−51.8 \times 10^{-5} .

Firstly, calculate the pKapK_a and pKbpK_b values.

  • pKa=−log⁡(Ka)=−log⁡(1.8×10−5)pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5})
  • pKb=−log⁡(Kb)=−log⁡(1.8×10−5)pK_b = -\log(K_b) = -\log(1.8 \times 10^{-5})

Given that the values of KaK_a and KbK_b are the same, their pKapK_a and pKbpK_b values will also be the same, establishing a neutral condition where the effects of the acid and base neutralize each other.

The formula used to determine the pH of a solution of such a salt is:

pH=12(pKw+pKa−pKb)

Where pKwpK_w is the ionic product of water, which is 14 at 25°C. In this specific case, since pKa=pKbpK_a = pK_b , the formula simplifies to:

pH=12(14+pKa−pKa)

pH=12(14)

pH=7

The pH of an ammonium acetate solution in this scenario is 7, indicating a neutral solution.

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