The ionic product of water

In a nutshell

$K_w$ stands for the ionic product of water. This is derived by the equilibrium constant and an assumption that the concentration of water is constant. It can be used to calculate the pH of strong bases. Additionally, $pK_a$ can be used to calculate the pH of weak acids.

Equations

word equation	symbol equation
$ionic\>product\>of\>water=hydrogen\>ions\>concentration\times\>hydroxide\>ions\>concentration$	$K_w =[H^+][OH^-]$
$pH=-\log_{10}{(hydrogen\>ions\>concentration)}$	$pH=-\log_{10}{[H^+]}$
$pK_w=-\log_{10}{(ionic\>product\>of\>water)}$	$pK_w=-\log_{10}{(K_w)}$
$pK_a=-log_{10}{(acid\>dissociation\>constant)}$	$pK_a=-log_{10}{(K_a)}$
$acid\>dissociation\>constant=10^{-pK_a}$	$K_a=10^{-pK_a}$
$acid\>dissociation\>constant=\dfrac{(hydrogen\>ions\>concentration)^2}{(weak\>acid\>concentration)}$	$K_a=\dfrac{[H^+]^2}{[HA]}$

Constants

constant	symbol	value
$ionic\>product\>of\>water\>for\>pure\>water$	$K_w$	$1.00\times10^{-14}\>mol^2\>dm^{-6}\>(at\>298\>K)$
$pK_w$	$pK_w$	$14.00\>(at\>298\>K)$

Variable units

quantity name	symbol	unit name	unit
$ionic\>product\>of\>water$	$K_w$	$mole\>squared\>per\>decimetre\>sixth$	$mol^2\>dm^{-6}$
$hydrogen\>ions\>concentration$	$[H^+]$	$mole\>per\>decimetre\>cubed$	$mol\>dm^{-3}$
$hydroxide\>ions\>concentration$	$[OH^-]$	$mole\>per\>decimetre\>cubed$	$mol\>dm^{-3}$
$acid\>dissociation\>constant$	$K_a$	$mole\>per\>decimetre\>cubed$	$mol\>dm^{-3}$
$weak\>acid\>concentration$	$[HA]$	$mole\>per\>decimetre\>cubed$	$mol\>dm^{-3}$

Ionic product of water, K_w

$K_w$ stands for the ionic product of water. This is derived by the equilibrium constant and an assumption that the concentration of water is constant. This is because the equilibrium lies to the left, there's not a lot of water dissociating. The concentration of water is much more than the concentration of hydrogen ions and hydroxide ions.

Example

Derive the ionic product of water using the equilibrium for water.

$H_2O \ (l)⇌H^+ \ (aq) +OH^- \ (aq)$

Write the $K_c$ equation.

$K_c = [H^+][OH^-]=K_w$

Pure water

Pure water is completely neutral. For pure water, the amount of hydrogen ions produced is equivalent to the amount of hydroxide ions produced. Therefore, their concentrations are equal. The ionic product of water for pure water can be written as:

$K_w=[H^+]^2$

Note: The value of $K_w$ at $298\>K$ is $1.00\times10^{-14}\>mol^2\>dm^{-6}$ .

K_w and strong base

The ionic product of water can be used to calculate the pH value of a strong base. Strong bases such as, $LiOH,\>NaOH$ and $KOH$ dissociate fully in water. This can be used to find the concentration of hydroxide ions as the concentration of the base will be considered the same as the concentration of the hydroxide ions.

The known concentration of hydroxide ions and $K_w$ can be used to calculate the concentration of hydrogen ions. Therefore, the pH can be calculated.

procedure

1.	Write the expression for $K_w$ . $K_w=[H^+][OH^-]$
2.	Rearrange for $[H^+]$ . $[H^+]=\dfrac{K_w}{[OH^-]}$
3.	Calculate $[H^+]$ .
4.	Write the equation for pH. $pH=-\log_{10}{[H^+]}$
5.	Calculate pH.

Example

Calculate the pH of $0.15\>mol\>dm^{-3}$ $KOH$ at $298\>K$ . The value of $K_w$ is $1.00\times10^{-14}\>mol^2\>dm^{-6}$ .

Write the expression for $K_w$ .

$K_w=[H^+][OH^-]$

Rearrange for $[H^+]$ .

$[H^+]=\dfrac{K_w}{[OH^-]}$

Identify $[OH^-]$ .

$[OH^-]=[strong\>base]=0.15\>mol\>dm^{-3}$ 

Insert known values to calculate $[H^+]$ .

 $[H^+]=\dfrac{1.00\times10^{-14}\>mol^2\>dm^{-6}}{{0.15}\>mol\>dm^{-3}}=6.666666667 \times10^{-14}\>mol\>dm^{-3}$

Write the pH equation.

$pH=-\log_{10}{[H^+]}$

Insert calculated $[H^+]$ value into pH equation to calculate pH.

$pH=-\log_{10}{[6.666666667 \times10^{-14}]}=13.18$

The pH is $\underline{13.18}$ .

Note: Always round to two decimal places for pH.

pK_w

The p stands for $-log_{10}$ . $K_w$ can be used to calculate $pK_w$ .

Example

Calculate $pK_w$ under standard conditions.

Write the expression for $pK_w$ .

$pK_w=-\log_{10}{K_w}$

The known value of $K_w$ under standard conditions is used.

$-\log_{10}{K_w}=-\log_{10}{1.00\times10^{-14}}=14.00$

The $pK_w$ is $\underline{14.00}$ .

pK_a and K_a

The acid dissociation constant, $K_a$ , is important to find the pH of a weak acid. Sometimes the acid dissociation constant needs to be calculated from $pK_a$ . The procedure to calculate pH using $pK_a$ is written below.

procedure

1.	Write the equation for $K_a$ from $pK_a$ . $K_a=10^{-pK_a}$
2.	Insert known $pK_a$ to calculate $K_a$ .
3.	Write the $K_a$ equation for calculation. $K_a=\dfrac{[H^+]^2}{[HA]}$
4.	Rearrange for $[H^+]$ . $[H^+]=\sqrt{K_a[HA]}$
5.	Insert known values to calculate $[H^+]$ .
6.	Write the pH equation. $pH=-\log_{10}{[H^+]}$
7.	Insert calculated $[H^+]$ to calculate pH.

Example

Calculate the pH of $0.030\>mol\>dm^{-3}$ ethanoic acid, $CH_3COOH$ . At this temperature, ethanoic acid has a $pK_a$ value is $3.20$ .

Write the equation linking $K_a$ and $pK_a$ .

$pK_a=-log_{10}{(K_a)}$

Rearrange for $K_a$ .

$K_a=10^{-pK_a}$

Insert known value to calculate $K_a$ .

$K_a=10^{-3.20}=6.309573445\times10^{-4}\>mol\>dm^{-3}$

Write the $K_a$ equation for calculating ethanoic acid.

$K_a=\dfrac{[H^+]^2}{[CH_3COOH]}$

Rearrange for $[H^+]$ .

$[H^+]=\sqrt{K_a[CH_3COOH]}$

Insert known values to calculate $[H^+]$ .

$[H^+]=\sqrt{6.309573445\times10^{-4}\times0.03}=\sqrt{1.892872033\times10^{-5}}=4.350714922\times10^{-3}\>mol\>dm^{-3}$

Write the pH equation.

$pH=-\log_{10}{[H^+]}$

Insert values to calculate pH.

 $pH=-\log_{10}{[4.350714922\times10^{-3}]}=2.36$

The pH is $\underline{2.36}$ .