Question

AH for the reaction H+(aq) + OH (aq) → H₂O is 56.8 kJ mol-¹. Given that the ionic product of water, Kw, is

0.61 x 10-¹4 at 291K, calculate K, and hence determine neutral pH at 310 K.

Answer 1

The equilibrium constant K for the reaction H+(aq) + OH-(aq) → H2O is 1.0 x
`10^{-14`. The neutral pH at 310 K is 7.

The ion product of water, Kw, is given by the equation Kw = [H3O+][OH-] and has a value of
`1.0 * 10^{-14` at 25°C. We can use this information to calculate K, the equilibrium constant, for the reaction H+(aq) + OH-(aq) → H2O. Since K = Kw at the same temperature, K = 1.0 x
`10^{-14`.

To determine the neutral pH at 310 K, we can use the equation

pH + pOH = 14.

The neutral pH of a solution is 7, which means the pOH is also 7. Therefore, pH = 14 - 7 = 7 at 310 K.