Question

What is the $\left[ {{H_3}{O^ + }} \right]$ for a solution with pH of $7.52$?

Answer 1

We have to know that the concentration of hydronium ion could be the same as the concentration of hydrogen ion. We can represent the concentration of hydronium ion as $\left[ {{H_3}{O^ + }} \right]$ and we can represent the concentration of hydrogen ion as $\left[ {{H^ + }} \right]$. We can calculate the pH of any substance with the help of the equation,
$pH = - \log \left[ {{H^ + }} \right]$

Complete step by step answer:
The given pH is $7.52$.
The negative log of the concentration of hydronium ion (or) negative log of the concentration of hydrogen ion is known as pH of the solution.
We can write the expression for calculating the pH of the solution as,
$pH = - \log \left[ {{H^ + }} \right]$
(or)
$pH = - \log \left[ {{H_3}{O^ + }} \right]$
Here, $\left[ {{H^ + }} \right]$ is indicated as the concentration of hydrogen ions.
$\left[ {{H_3}{O^ + }} \right]$ is indicated as the concentration of hydronium ions.
When we rearrange the expression of pH of the solution, we will get the concentration of hydronium ion.
We can rearrange the expression to get concentration of hydronium ion as,
$pH = - \log \left[ {{H_3}{O^ + }} \right]$
$\left[ {{H_3}{O^ + }} \right] = {10^{ - pH}}$
We can calculate the concentration of hydronium ion as,
$\left[ {{H_3}{O^ + }} \right] = {10^{ - pH}}$
$\left[ {{H_3}{O^ + }} \right] = {10^{ - 7.25}}$
$\left[ {{H_3}{O^ + }} \right] = 5.6234M$
We have calculated the concentration of hydronium ion as $5.6234M$.
Note: We have to know that for base, we have to determine the $pH$ with the help of the $pOH$ of the solution. When it comes to acid, we have to determine the $pH$ with the help of the concentration of ${H^ + }$. So, when determining the concentration which is nothing but the molarity, we have to determine the moles (if grams are given) and volume. When the volume is given in terms of milliliters, we should convert milliliters to liters. When the conversion of milliliters to liters is not done, the correct value of $pH$ could not be obtained.