Question

At ${100^0}C$,${K_w} = {10^{ - 12}}$. The $pH$ Of pure water at ${100^0}C$ will be:
A.$7.0$
B.$6.0$
C.$8.0$
D.$12.0$

Answer 1

Hint: To solve the given problem we must be aware of the basics of the ionic equilibrium. In this question, we need to find $pH$ Of pure water so we will need the concentration of $[{H^ + }]$ ions. We can get the concentration using the condition given in the question.
Formula Used:
The formula of $pH$
$pH = - \log [{H^ + }]$
Where $pH$ represents the potential of hydrogen,
$[{H^ + }]$ The concentration of hydrogen ions.
${K_w} = [{H^ + }][O{H^ - }]$
Where ${K_w}$ represents the ionization constant of water,
$[{H^ + }]$ The concentration of hydrogen ion,
$[O{H^ - }]$ The concentration of hydroxyl ions.

Complete step by step answer:
First, we will understand the terms given in the question. So, first, we will define $pH$. $pH$ is defined as the measure of the relative amount of hydrogen and hydroxyl ions in the water. We can also define it numerically as the negative logarithm of hydrogen ion concentration.
In the question, we have one more term ${K_w}$. It is the equilibrium constant also called the dissociation constant or ionization constant of water.
Now we have the basic idea of the quantities given in the question. So we are all set to solve the numerical.
Now we will write the given quantities from the question. We have the value of ${K_w}$. So we can write it as ${K_w} = {10^{ - 12}}$. We can write it as,
${K_w} = [{H^ + }][O{H^ - }] = {10^{ - 12}}$ $ - \left( 1 \right)$
Now in the question, we have given pure water and we know that for pure water the concentration of hydrogen ion is equal to the concentration of hydroxyl ion. So using the equation $\left( 1 \right)$ and the condition of pure water we can conclude that,
$[{H^ + }] = [O{H^ - }] = {10^{ - 6}}$
Now we have the concentration of hydrogen ion as $[{H^ + }] = {10^{ - 6}}$. We can find $pH$ pure water by substituting the value of hydrogen ion concentration in the formula.
$pH = - \log [{H^ + }]$
$ \Rightarrow pH = - \log [{10^{ - 6}}]$
Now using the property of logarithm, $\log ({a^b}) = b\log (a)$ we get,
$pH = - ( - 6)\log [10]$
$ \Rightarrow pH = 6.0$ $\left( {\log (10) = 1} \right)$
$pH$ Of pure water at ${100^0}C$ will be $pH = 6.0$.
Hence, the correct option is (B).

Note:
-The $pH$ is an important quantity as it reflects the chemical condition of a solution. The value $pH$ ranges from $0 - 14$ .
-The $pH$ water is important as it keeps our body in balance and maintains the metabolic processes.