Question

Calculate the pH of a solution with [H+] = 5.6 x 10^-8

Answer 1

Answer:

Approximately $7.25$.

Explanation:

If the concentration of ${\rm H^{+}}$ ions in a solution is $x\; {\rm M}$, the ${\rm pH}$ of that solution would be $(-\log_{10}(x))$.

Note that the base of the logarithm in this calculation should $10$. One way to be sure is to state the base explicitly. Using the change of base rule of logarithms:

$\begin{aligned}-\log_{10}(x) = -\frac{\ln(x)}{\ln(10)}\end{aligned}$.

In this question, it is implied that the concentration of ${\rm H^{+}}$ in the given solution is $5.6 \times 10^{-8}\; {\rm M}$, such that $x = 5.6 \times 10^{-8}$. Using the equations above:

$\begin{aligned}& \text{pH of this solution} \\ =\; & -\log_{10}(x) \\ =\; & -\frac{\ln(x)}{\ln(10)} \\ =\; & -\frac{\ln(5.6 \times 10^{-8})}{\ln(10)} \\ \approx\; & 7.25\end{aligned}$.