# How do we work out [H_3O^+] from pH?

$\left[{H}_{3} {O}^{+}\right] = {10}^{- p H}$.
Back in the day, we had to use logarithmic tables, and take the antilogarithm. These days, given a calculator, put in the $p H$ value, hit the $\text{+/-}$ button to make this value negative, and then hit the button ${10}^{x}$, to give ${10}^{- p H}$.
Most calculators have the $\log$ button on a shift with ${10}^{x}$. Anyway, if I have not answered your question, someone will try again.