# Question 4e291

May 13, 2016

["H"_3"O"^(+)] = 2.24 * 10^(-5)"M"

#### Explanation:

Hydrochloric acid, $\text{HCl}$, is a strong acid that ionizes completely in aqueous solution to form hydronium cations, ${\text{H"_3"O}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$.

As you can see, every mole of hydrochloric acid will produce one mole of hydronium cations and one mole of chloride anions

The pH of the solution is calculated by taking the negative log base $10$ of the concentration of hydronium cations

color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))

You can express the concentration of the hydronium cations in term of the solution's pH by rewriting the above equation as

log(["H"_3"O"^(+)]) = -"pH"

This will be equivalent to

10^log(["H"_3"O"^(+)]) = 10^(-"pH")

which will give you

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \left[\text{H"_3"O"^(+)] = 10^(-"pH}\right) \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, the solution is said to have a pH of $4.65$. This corresponds to a concentration of hydronium cations equal to

$\left[{\text{H"_3"O}}^{+}\right] = {10}^{- 4.65}$

["H"_3"O"^(+)] = color(green)(|bar(ul(color(white)(a/a)2.24 * 10^(-5)"M"color(white)(a/a)|)))#