Question

1. For the autoionization of water.

(For this reaction, H° = 5.58 x 104 J)

b) Calculate [H3O+] at (i) 25 °C and (ii) at 100 °C for pure water.

Answer 1

2 H₂O ⇄ H₃O⁺ + OH⁻

The equilibrium equation for this reaction is:
Kw = [H₃O⁺]∙[OH⁻]

According to reaction same amounts of H₃O⁺ and OH⁻ are formed. So the molar concentration two ionic species in an equilibrium mixture is the same:
[H₃O⁺] = [OH⁻]
Hence,
Kw = [H₃O⁺]²
=>
[H₃O⁺] = √Kw

The given value for Kw is the value at standard temperature of 25°C. The value of Kw at 100°C can be found using the van't hoff equation [1]. It relates the the values of the equilibrium constant at two distinct temperatures as:
ln(K₂/K₁) = (ΔH/R)∙((1/T₁) - (1/T₂))
=>
K₂ = K₁ ∙ e^{ (ΔH/R)∙((1/T₁) - (1/T₂)) }

So autoionizatiion constant at 100°C is
Kw = 1.0×10⁻¹⁴ ∙ e^{ (5.58×10⁴ J∙mol⁻¹ / 8.3145 J∙K⁴∙mol⁻¹)∙((1/(298 K) - (1/373 K)) }
= 9.26×10⁻¹³

The concentration of hydronium ions is:

(i) at 25°C
[H₃O⁺] = √1.0×10⁻¹⁴ = 1.0×10⁻⁷ M

(i) at 100°C
[H₃O⁺] = √9.26×10⁻¹³ = 9.62×10⁻⁷ M