# A 10^-3 M solution has been diluted to 100 times.Calculate the pH of the diluted solution?

May 19, 2016

#### Explanation:

Since you didn't specify the content of your solution, I will have to assume that you're dealing with a solution that contains a strong monoprotic acid like hydrochloric acid, $\text{HCl}$.

Hydrochloric acid ionizes in a $1 : 1$ mole ratio to form hydronium cations, ${\text{H"_3"O}}^{+}$, which means that your initial solution will have

["H"_3"O"^(+)] = 10^(-3)"M"

Now, you dilute this solution by a factor of $100$, which essentially means that the volume of the final solution will be $100$ times bigger than the volume of the initial solution.

This means that if you take $V$ to be volume of your starting solution, and $n$ the number of moles of hydronium cations that it contains, you will have

["H"_3"O"^(+)] = n/V = 10^(-3)"M"-> for the initial solution

After the initial solution is diluted, its volume will be $100 \cdot V$. You can thus say that

$\left[{\text{H"_3"O}}^{+}\right] = \frac{n}{100 \cdot V} = \frac{1}{100} \cdot \frac{n}{V}$

["H"_3"O"^(+)] = 1/100 * 10^(-3)"M" = 10^(-5)"M"

Now ll you have to do is use the equation

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

to find the pH of the dilutes solution. In your case, you will have

$\text{pH} = - \log \left({10}^{- 5}\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 5 \textcolor{w h i t e}{\frac{a}{a}} |}}}$