Q1(h). log k against pH for specific H⁺ catalysis

Answer in brief

A straight line with slope = −1 and intercept (at pH 0) = log kH⁺.

Key points

• kobs = kH⁺[H⁺] ⇒ log kobs = log kH⁺ + log[H⁺] ⇒ log kobs = log kH⁺ − pH.
• Specific acid catalysis: only H₃O⁺ catalyses; the rate is first order in [H⁺].
• Slope = −1. Intercept = log kH⁺, where kH⁺ is the catalytic coefficient (kobs at [H⁺] = 1 M).
• If an uncatalysed path exists, kobs = k₀ + kH⁺[H⁺] and the line flattens at high pH.