Question

The \[\left[ {{H^ + }} \right]\](${H^ + }$concentration) in a solution is \[1 \times {10^{ - 8}}M\]. What is the $pH$ of the solution?

Answer 1

Hint: We need to know the concept of $pH$ and accordingly calculate the $pH$ with the given hydrogen concentration. We know that $pH$ is defined as the negative log of hydrogen ion concentration but in terms of molarity, a solution with $pH$ $1$ has \[{10^{ - 1}}M\] of hydrogen ion concentration and that of $pH$ \[2\] would be \[{10^{ - 2}}M\] of hydrogen ion concentration. Since the $pH$is increasing and molarity is decreasing, we can call it dilution.
Formula used:
\[pH{\text{ }} = {\text{ }} - log\left[ {{H^ + }} \right]\]

Complete answer:
$pH$ is a chemistry scale for determining the acidity or basicity of aqueous solutions. Acidic solutions (those with a greater concentration of ${H^ + }$ ions) have lower $pH$ than basic or alkaline solutions. The $pH$ scale is logarithmic, indicating the concentration of hydrogen ions in a solution in reciprocal order. Since the $pH$ formula approximates the negative of the base $10$ logarithm of the molar concentration of hydrogen ions in the solution, this is the case. $pH$ is defined as the negative of the base $10$ logarithm of the ${H^ + }$ion's action.
Given that the ${H^ + }$concentration in a solution is \[1 \times {10^{ - 8}}M\]. Using the formula \[pH{\text{ }} = {\text{ }} - log\left[ {{H^ + }} \right]\], we can calculate $pH$of the solution.
\[pH{\text{ }} = {\text{ }} - log\left[ {1 \times {{10}^{ - 8}}M} \right]\]= $8$
Therefore, the $pH$ of the solution is $8$.

Note:
It must be noted that the concentration of a given compound can be considered to be the concentration of their dissociated ions in the case of strong acids and bases only. In the case of weak acids or weak bases where they do not dissociate completely, the concentration of the acid or the base will be more than its dissociated ions.