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 $