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

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, #"HCl"#.

Hydrochloric acid ionizes in a #1:1# mole ratio to form hydronium cations, #"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 * V#. You can thus say that

#["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)|)))#