Question

What is the $pH$ of pure water at $25^0C$ if the $Kw$ at this temperature is $1.0 \times 10^{ -14 }$ ?

Answer 1

Hint : To solve the given problem, we should have information about $pH$ and $Kw$ . Basically determines the power of hydronium ions. It is also the potential of hydrogen. It is inversely proportional of hydrogen. It is inversely proportional to the concentration of $H^+$ ion.
$Kw$ or the autoionisation constant of water is the equilibrium constant when water dissociates or associates. It is determined during autolysis.
First we take a look into autolysis and determine $Kw$ as $Kw =[H_3O^+] [OH^-]$ .
And second formula is $pH = -log[H^+]$
Where, $[H^+] \rightarrow$ Concentration of $H^+$ ion.

Complete Step By Step Answer:
Step-1 :
At $25^0C$ temperature, i.e. at room temperature, autolysis of water occur as :
$H_2O+H_2O \Leftrightarrow H_3O^+OH^-$
Here, equilibrium constant will be :
$K_a = dfrac{[H_3O^+][OH^-]}{[H_2O]^2 =[H_3O^+][OH^-]}$
It is also referred to as $Kw$ .
Step-2:
In case of pure water as it is given in the question,
$[H_3O^+] = [OH^-]$
So, $Kw = [H_3O^+][OH^-]$
$Kw = [H_3O^+]^2$
We know, $Kw = 1 \times 10^{ -14 }$
So, $[H_3O^+]^2 = 10^{ -14 }$
$[H_3O^+] = 10^{ -7 }$
Step-3 :
Using the given formula, we have :
$pH = -log[H^+]$ Or, $pH = -log[H_3O^+]$
$pH = -log[10^{ -7 }]$
Using rule $loga^m = mloga$ , we have ;
$pH = 7log10$
So, $pH = 7$

Note :
The reactants can be of different types in neutralization reactions like strong acid and strong base, weak acid and strong base or vice versa. The $pH$ change in any reaction basically depends upon the relative strength of the reactants.
In case of neutralization, the $pH$ is always nearly $7$ or equal to $7$ .
The $pH$ for acids is less than seven. The lesser the value, the stronger is the acid. The $pH$ for base is greater than seven. The greater the value, the stronger the base. The $pH$ range is from $0-14$ .
The formula for $pH$ is ;
$pH = -log[H^+]$
The formula for $pOH$ is ;
$pOH = -log[OH^-]$
And,
$pH + pOH = 14$