## Autoionization Of Water

The autoionization of water is an equilibrium reaction that is written:

H

_{2}O(l) H^{+}(aq) + OH^{-}(aq)The equilibrium constant expression for this reaction is given by:

K

_{eq}= K_{w}= [H^{+}] [OH^{-}] = 1.0 x 10^{-14}The value for K

_{w}is for room temperature, 25 °C, and 1.0 atm of pressure.The equilibrium constant expression applies not only to pure (distilled) water but to any aqueous solution. It can be used to calculate either [H

^{+}] or [OH^{-}] provided one of them is known.When [H

^{+}] = [OH^{-}], the solution is neutral, when the [H^{+}] > [OH^{-}] the solution is acidic, and when [H^{+}] < [OH^{-}] the solution is basic. [H^{+}] and [OH^{-}] are inversely proportional meaning their product must always equal the constant, 1.0 x 10^{-14}, at 25 °C, and 1.0 atm of pressure.The degree of ionization and the value of K

_{w}varies with temperature as shown in the table below.

Temp

(°C)

1/Temp

(x 10

^{-3}K^{-1})K

_{w} ln K_{w}pK

_{w}0

3.661.14 x 10

^{-15} -34.4114.9435

5

3.601.85 x 10

^{-15} -33.9214.7338

10

3.532.92 x 10

^{-15} -33.4714.5346

15

3.474.53 x 10

^{-15} -33.0314.3436

20

3.416.81 x 10

^{-15} -32.6214.1669

25

3.361.01 x 10

^{-14} -32.2313.9965

30

3.301.47 x 10

^{-14} -31.8513.8330

35

3.252.09 x 10

^{-14} -31.5013.6801

40

3.192.92 x 10

^{-14} -31.1613.5348

45

3.144.02 x 10

^{-14} -30.8413.3960

50

3.105.47 x 10

^{-14} -30.5413.2617

The form of the van't Hoff equation which is valid for small temperature ranges is given by:

ln K = -ΔH

_{rxn}°/RT + ΔS_{rxn}°/RComparing this equation to y = mx + b, suggests that a plot of ln K

_{w}vs 1/T will be a straight line with a slope of -ΔH_{rxn}°/R and the term ΔS_{r}_{xn}°/R is the y-intercept as shown in the graph below.

Applying the van't Hoff equation and ignoring the y-intercept gives:

ln (K

_{2}/K_{1}) = (-ΔH_{rxn}°/R)(1/T_{2}- 1/T_{1})Substituting the values corresponding to 50 °C and 0 °C gives:

ln ((5.47 x 10

^{-14})_{}/(1.14 x 10^{-15})_{}) = (-ΔH_{rxn}°/8.31 J/mol•~~K~~)(3.10 x 10^{-3}~~K~~- 3.66 x 10^{-3}~~K~~)ΔH

_{rxn}° = 57.43 kJ/molThe graph below shows the line of best fit (linear regression) for the data above.

The graph of pK

_{w}vs 1/T shown below also gives the same result. The slope, m, of the graph below is given by:m = ΔpK

_{w}/Δ(1/T) = pK_{w2}- pK_{w1}/(1/T_{2}- 1/T_{1})The slope of the graph, m, also equals:

m = ΔH

_{rxn}°/(2.3026•R)Equating these two expressions for m and substituting values yields:

(13.262 - 14.944)/(323

^{-1}~~K~~- 273^{-1}^{-1}~~K~~) = ΔH^{-1}_{rxn}°/(2.3026•8.31~~J~~/mol•~~K~~•1 kJ/10^{3}~~J~~)ΔH

_{rxn}° = 56.76 kJ/mol

The graph below shows the line of best fit (linear regression) for the data above.

The value for ΔH

_{rxn}° obtained from the graph compares favorably to ΔH_{rxn}° when computed using enthalpies.H

_{2}O(l) H^{+}(aq) + OH^{-}(aq)ΔH

_{rxn}° =ΣnΔH_{f}°(products) -ΣmΔH_{f}°(reactants)ΔH

_{rxn}° =[ΔH_{f}°(H^{+}(aq)) + ΔH_{f}°(OH^{-}(aq))] - [ΔH_{f}°(H_{2}O)]ΔH

_{rxn}° =(0 + 1~~mol OH~~x -230.0 kJ/^{-}~~mol OH~~) - (1^{-}~~mol H~~x -285.83 kJ/_{2}O~~mol H~~) = 55.8 kJ_{2}O