Question

How do you calculate the pH of HCl?

Answer 1

Hint:. First we should know about the concentration of HCl solution or the molarity of the solution, then by applying the pH formula as $pH=-\log [{{H}^{+}}]$, we can easily find the pH of the solution. Now solve it.

Complete step by step answer:
- First of all, let’s discuss what is pH. pH gives us the measure of acid/base strength of any solution. pH scale ranges from 0-14, 7 is neutral, below 7 it represents acidic nature i.e. the compound is acidic and above 7 it represents basic nature i.e. the compound is basic.
- HCl stands for hydrochloric acid and it is a very strong acid. So, from this we come to know that its pH should fall in the range between $0-7$.
Now, considering the statement:
- First, of all we should know about the molarity of the solution i.e. the concentration of HCl in $mol\text{ lite}{{\text{r}}^{-1}}$ and then we can easily calculate the pH of HCl solution by using the pH formula as;
$pH=-\log [{{H}^{+}}]$ -----------(1)
i.e. pH is the negative logarithm of hydrogen ion.
- Here, ${{H}^{+}}$ represents the concentration of the hydrogen ion in the solution and the concentration of both the acid i.e. HCl and hydrogen ions i.e. ${{H}^{+}}$ in the solution is the same. It is so because, HCl is a very strong acid and dissociates completely into the hydrogen ions and chloride ions as;
 $HCl\to {{H}^{+}}+C{{l}^{-}}$
- Now, if suppose that the molarity of solution is $0.1M$
Then, the pH of the HCl solution is as;
Using equation (1), we get:
$\begin{align}
  & pH=-\log [0.1] \\
 & \text{ = }-\log [1\times {{10}^{-1}}] \\
 & \text{ = }-\log [{{10}^{-1}}] \\
 & \text{ =}-(-1)\log 10\text{ (log10=1)} \\
 & \text{ =1} \\
\end{align}$
Thus, in this way we can find the pH of the HCl solution.

Note: Molarity may be defined as the no. of moles of the solute to the total volume of the solution in liters or $1000\text{ ml}$ of the solution. it is represented by the symbol as M.
$molarity=\dfrac{no.\text{ }of\text{ }moles\text{ }of\text{ }the\text{ }solute}{total\text{ }volume\text{ }of\text{ }solution\text{ }in\text{ }liters}$