The pH of pure water at 298 K and 308 K are 7 and 6 respectively. Calculate the heat of formation of water from H + and O H – .

Step-by-step solution

At 298 K; [ H + ] = 10 – 7 ∴ K w 1 = 10 − 14 At 308 K; H + = 10 – 6 ∴ K w 2 = 10 − 12 Now using 2.303 log 10 Kw 2 Kw 1 = ΔH R T 2 − T 1 T 1 × T 2 2.303   log 10 10 − 12 10 − 14 = ΔH 2 308 − 298 298 × 308 ∴ Δ H = 84551.4 k c a l / m o l = 84.551 k c a l / m o l Thus H 2 O ⇌ H + + O H - ; Δ H = 84.551 k c a l / m o l ∴ H + + O H - ⇌ H 2 O ; Δ H - 84.551 k c a l / m o l