Question

Values of Kw as a function of temperature are as follows:

0	1.14*10^-15
25	1.00*10^-14
35	2.09*10^-14
40	2.92*10^-14
50	5.47*10^-14

Is the auto ionization of water exothermic or endothermic? Explain

. What is the pH of water at 50ºC?

Determine Kw at 37ºC, normal physiological temperature.

Determine the pH of a neutral solution at 37ºC (using an Arrhenius plot of the data).

What is the bond dissociation enthalpy for breaking one O—H bond in water to form H+ and OH–? Compare this calculated value to that found in bond dissociation enthalpy table (is this the identical reaction? Write the (comparable) reaction. Explain)

Answer 1

The plot of ln(Kw) versus 1/T looks like:

The equation is shown on graph.

Slope = -6183.2

So, - Ea/R = - 6183.2

Or, Ea = + 6183.2 * R

So, the energy change is positive and the autoionisation of water is endothermic.

At 50 ⁰C, Kw = 5.47*10^-14

So, [H⁺] = (Kw)^1/2 = ( 5.47*10^-14 )^1/2 = 2.34 * 10^-7

So, pH = - log [H+] = - log ( 2.34 * 10^-7 ) = 6.64

From the plot we get: Slope = 6183.2 * R and intercept is : -11.454

when T = 37 ⁰C = 310 K ,

ln (Kw) = - 6183.2 * R/( R * 310)   - 11.454

Or, ln Kw = -31.4

So, Kw = e^-31.4 = 2.31 * 10^-14

kw at 37 ⁰C is 2.31 * 10^-14

So, [H+] = ( 2.31 * 10^-14 ) ^1/2 = 1.52 * 10^-7

pH = - log [H+] = - log ( 1.52 * 10^-7 ) = 6.81