A phenol derivative has an acid dissociation constant K_a = 2.5×10^-10 at 25°C. Calculate its pK_a.

A phenol derivative has an acid dissociation constant $K_{a} = 2.5 \times 1 0^{- 10}$ at $2 5^{\circ} C$ . Calculate its $p K_{a}$ .

easy 1 mark Class 12 · Chemistry · Alcohols Phenols and Ethers

Answer

p K_{a} = 9.60

The correct answer is $p K_{a} = 9.60$ — Step 5: Round to two decimal places: $p K_{a} a pp r o x 9.60$ .

Step 1: Write the relation $p K_{a} = - l o g_{10} K_{a}$ .
Step 2: Substitute $K_{a} = 2.5 \times 1 0^{- 10}$ : $p K_{a} = - l o g_{10} (2.5 \times 1 0^{- 10})$ .
Step 3: Split the logarithm: $l o g_{10} (2.5 \times 1 0^{- 10}) = l o g_{10} 2.5 + l o g_{10} 1 0^{- 10} = 0.3979 - 10 = - 9.6021$ .
Step 4: Apply the negative sign: $p K_{a} = - (- 9.6021) = 9.6021$ .
Step 5: Round to two decimal places: $p K_{a} a pp r o x 9.60$ .

Common mistake: Forgetting the negative sign in the logarithm conversion

Concept tested: pKa calculation using Ka

Published 11 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked