Question

Calculate $pH$ of $0.0001\;M$ of $HN{O_3}$.

Answer 1

Hint: The compound given in the question is Nitric acid, it is a very strong acid and a very strong oxidizing agent. It is highly corrosive and toxic in nature, if it comes in contact it can burn the skin. It is colourless but with time it turns into yellow colour because of decomposition of nitric acid into oxides of nitrogen and water.

Complete answer:
It contains three Oxygen atoms and one Hydrogen atom and one Nitrogen atom. We will calculate the concentration of Hydrogen ions in the solution because $pH$ is the measure of ${{\text{H}}^ + }$ ions in the solution. It is calculated by the formula
$pH\; = \; - {\log _{10}}\left[ {{{\text{H}}^ + }} \right]$
Let’s look at the reaction
$HN{O_3}\; \rightleftarrows \;{{\text{H}}^ + }\; + \;N{O_3}^ - $
Nitric acid dissociated into Hydrogen ion and Nitrate ion the concentration of both ions is $0.0001\;M$ so let us calculate the $pH$of the given amount of nitric acid.
$pH\; = \; - {\log _{10}}\left[ {{{\text{H}}^ + }} \right]$
substituting the value of concentration of hydrogen ions in the nitric acid, we get
$pH\; = \; - {\log _{10}}\;\left[ {0.0001} \right]$
simplifying the equation, we get
$pH\; = \; - {\log _{10}}\;\left[ {{{10}^{ - 4}}} \right]$
as we know ${\log _{10}}10\; = \;1$ so ${\log _{10}}\left[ {{{10}^{ - 4}}} \right]\; = \; - 4$
$pH\; = \; - \left( { - 4} \right)$
$pH\; = \;4$
so, the $pH$ of the given amount of nitric acid is four.

Note:
Be careful when doing the calculations of the $pH$ as it involves the calculation of $\log $. we write in the from of ${\log _a}B\; = \;c$ where ${{\text{a}}^{\text{c}}}\; = \;{\text{B}}$ pay attention to the $\log $ part as it is necessary for the calculation of the $pH$ of the solution.