# Question fdeda

May 2, 2016

Here's how you can do that.

#### Explanation:

Let's assume that you're dealing with a monoprotic strong acid like hydrochloric acid, $\text{HCl}$.

Strong acids ionize completely in aqueous solution to form hydronium cations, ${\text{H"_3"O}}^{+}$, in a $1 : 1$ mole ratio

color(red)("H")"Cl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(color(red)(+)) + "Cl"_((aq))^(-)

This means that every mole of strong acid will produce one mole of hydronium cations in solution. Therefore, for a strong monoprotic acid solution, you have

$\left[\text{H"_3"O"^(+)] = ["concentration of the acid}\right]$

["HCl"] = "0.0001 mol L"^(-1)

This means that the solution will contain

["H"_3"O"^(+)] = "0.0001 mol L"^(-1)

Since the pH of a solution is defined as 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 say that the pH of this solution will be

$\text{pH} = - \log \left(0.0001\right) = - \log \left({10}^{- 4}\right) = 4 + \log \left(1\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 4 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, an interesting example would be sulfuric acid, ${\text{H"_2"SO}}_{4}$, which is considered, for almost all intended purposes, a diprotic strong acid.

When sulfuric acid ionizes in water, it releases two protons, which means that it produces hydronium cations in a $1 : 2$ mole ratio.

color(red)("H")_ color(green)(2)"SO"_ (4(aq)) + 2"H"_ 2"O"_ ((l)) -> color(green)(2)"H"_ 3"O"_ ((aq))^(color(red)(+)) + "SO"_(4(aq)^(-)

This time, you have

$\left[\text{H"_3"O"^(+)] = 2 xx ["concentration of the acid}\right]$

In your case, you will end up with

["H"_3"O"^(+)] = 2 xx "0.0001 mol L"^(-1) = "0.0002 mol L"^(-1)#

The pH will be

$\text{pH} = - \log \left(0.0002\right) = - \log \left(2 \cdot {10}^{- 4}\right) = 4 + \log \left(2\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 3.7 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The pH of this solution is lower than the pH of the first solution because you have a higher concentration of hydronium cations.