Wize High School Grade 12 Chemistry Textbook > Acids and Bases

Autoionization of Water (Kw)

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  Autoionization of Water (Kw)
Two water molecules can react with themselves in an autoionization reaction as follows: 

2H2O(l)↔H3O(aq)++OH(aq)−2H_2O_{(l)}\leftrightarrow H_3O_{(aq)}^++OH_{(aq)}^-2H2​O(l)​↔H3​O(aq)+​+OH(aq)−​

Kw=[H3O(aq)+][OH(aq)−]=1×10−14 (25oC)\boxed{K_w=[H_3O_{(aq)}^+][OH_{(aq)}^-]=1\times10^{-14}\ (25^oC)}Kw​=[H3​O(aq)+​][OH(aq)−​]=1×10−14 (25oC)​
Here is the reaction in action!

Photo by Rice University / CC BY

Label whether each reactant and product in the above reaction is acting as an acid/base or conjugate acid/conjugate base. 

Water is amphoteric: it can act as an acid or a base! 

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Other Key Relationships
Kw = [H+][OH-] = 1x10-14,  [H+]= 1x10-7M and [OH-]= 1x10-7M at 25oC

This is why the pH (and pOH) of neutral water is 7 

Kw=[H3O(aq)+][OH(aq)−]=(Ka)(Kb)=1×10−14 (25oC)\boxed{K_w=[H_3O_{(aq)}^+][OH_{(aq)}^-]=(Ka)(Kb)=1\times10^{-14}\ (25^oC)}Kw​=[H3​O(aq)+​][OH(aq)−​]=(Ka)(Kb)=1×10−14 (25oC)​
 *Ka and Kb have to be the K values for an acid and its conjugate base (or a base and its conjugate acid)

If [H+] goes higher, what do you think would happen to [OH-]? Goes higher/lower: lower

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Calculate [H+], [OH-], pH, and pOH for the following:

a) pH=3

[H+]=10-pH
[H+]=10-3
[H+]=1x10-3M

Kw=[H+][OH-]

Kw[H+]=[OH−]\frac{Kw}{\left[H+\right]}=\left[OH-\right][H+]Kw​=[OH−]

(1x10−14)(1x10−3)=[OH−]\frac{\left(1x10^{-14}\right)}{\left(1x10^{-3}\right)}=\left[OH^-\right](1x10−3)(1x10−14)​=[OH−]

[OH-]=1x10-11M 

pH + pOH =14
14-3=pOH
pOH=11

Could get the same answer by using pOH=-log[OH-]

b) pOH=5.63

[OH-]=10-pOH
[OH-]=10-5.63
[OH]=2.34x10-6

Kw=[H+][OH-]

Kw[OH−]=[H+]\frac{Kw}{\left[OH-\right]}=\left[H+\right][OH−]Kw​=[H+]

[H+]=1x10−142.34x10−6\left[H+\right]=\frac{1x10^{-14}}{2.34x10^{-6}}[H+]=2.34x10−61x10−14​

[H+]=4.27x10-9

pH +pOH =14
14-pOH=pH
14-5.63=pH
pH=8.37
Could get same answer by using pH=-log[H+]

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Example: pH Problem  
C6H5COOH    ⇌      H+ + C6H5COO-  

Ka of C6H5COOH = 6.5 x 10-7 and the concentration of C6H5COOH is 0.01 M, what is the pH of the solution?

Write Out An ICE Table
I  0.01M    0        0
C -x           +x      +x
E 0.01-x    x        x

Write a K expression and Solve for x
Ka = x²/(0.01 - x)
Ka = x²/0.01
We can ignore the "-x" part since y/K > 400

x² = Ka·0.01
x² = (6.5 x 10-7)(0.01)
x² = 6.5 x 10-9
x = 8.06 x 10-5
x=[H+]

Find pH
pH = -log[H+]
pH = -log(8.06 x 10-5)
pH = 4.09