Autoionization Of Water

 

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

H2O(l) H+(aq) + OH-(aq)

The equilibrium constant expression for this reaction is given by:

Keq = Kw = [H+] [OH-] = 1.0 x 10-14

The value for Kw 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 Kw varies with temperature as shown in the table below.

 

Temp

(°C)

1/Temp

 (x 10-3 K-1)

Kw

ln Kw

pKw

0

3.66

1.14 x 10-15

-34.41

14.9435

5

3.60

1.85 x 10-15

-33.92

14.7338

10

3.53

2.92 x 10-15

-33.47

14.5346

15

3.47

4.53 x 10-15

-33.03

14.3436

20

3.41

6.81 x 10-15

-32.62

14.1669

25

3.36

1.01 x 10-14

-32.23

13.9965

30

3.30

1.47 x 10-14

-31.85

13.8330

35

3.25

2.09 x 10-14

-31.50

13.6801

40

3.19

2.92 x 10-14

-31.16

13.5348

45

3.14

4.02 x 10-14

-30.84

13.3960

50

3.10

5.47 x 10-14

-30.54

13.2617

 

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

ln K = -ΔHrxn°/RT + ΔSrxn°/R

Comparing this equation to y = mx + b, suggests that a plot of ln Kw vs 1/T will be a straight line with a slope of -ΔHrxn°/R and the term ΔSrxn°/R is the y-intercept as shown in the graph below.

 

 

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

ln (K2/K1) = (-ΔHrxn°/R)(1/T2 - 1/T1)

Substituting the values corresponding to 50 °C and 0 °C gives:

ln ((5.47 x 10-14)/(1.14 x 10-15)) = (-ΔHrxn°/8.31 J/mol•K)(3.10 x 10-3 K - 3.66 x 10-3 K)

ΔHrxn° = 57.43 kJ/mol

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

 

 

The graph of pKw vs 1/T shown below also gives the same result. The slope, m, of the graph below is given by:

m = ΔpKw/Δ(1/T) = pKw2 - pKw1/(1/T2 - 1/T1)

The slope of the graph, m, also equals:

m = ΔHrxn°/(2.3026R)

Equating these two expressions for m and substituting values yields:

(13.262 - 14.944)/(323-1 K-1 - 273-1 K-1) = ΔHrxn°/(2.30268.31 J/mol•K•1 kJ/103 J)

ΔHrxn° = 56.76 kJ/mol

 

 

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

 

 

The value for ΔHrxn° obtained from the graph compares favorably to ΔHrxn° when computed using enthalpies.

H2O(l) H+(aq) + OH-(aq)

ΔHrxn° = ΣnΔHf°(products) - ΣmΔHf°(reactants)

ΔHrxn° = [ΔHf°(H+(aq)) + ΔHf°(OH-(aq))] - [ΔHf°(H2O)]

ΔHrxn° = (0 + 1 mol OH- x -230.0 kJ/mol OH-) - (1 mol H2O x -285.83 kJ/mol H2O) = 55.8 kJ