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

##### 1 Answer

Here's one possible answer.

#### 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,

Hydrochloric acid ionizes in a **mole ratio** to form hydronium cations,

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

Now, you dilute this solution by a factor of *final solution* will be **times** bigger than the volume of the initial solution.

This means that if you take *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

#["H"_3"O"^(+)] = n/(100 * V) = 1/100 * 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

#"pH" = - log(10^(-5)) = color(green)(|bar(ul(color(white)(a/a)5color(white)(a/a)|)))#