# Question #afe36

##### 1 Answer

Here's what I got.

#### Explanation:

The idea here is that you're essentially performing a dilution, so you don't *really* need to know how many moles of hydronium ions,

Also, the problem tells you that you're dealing with a **strong acid**, so you won't be using any

Like you said, the first step is to calculate the molarity of the hydronium ions in the initial solution. Since

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

you get that

#["H"_3"O"^(+)]_1 = 10^(-"pH"_1)#

#["H"_3"O"^(+)]_1 = 10^(-0.5) = "0.3162 M"#

Now use the pH of the target solution to find out what concentration of hydronium ions would produce a pH of

#["H"_3"O"^(+)]_2 = 10^(-"pH"_2)#

#["H"_3"O"^(+)]_2 = 10^(-0.7) = "0.1995 M"#

Now, when you're **diluting** a solution, you're essentially keeping the number of moles of solute **constant**.

The concentration of the target solution is **lower** than the concentration of the initial solution because you're **adding solvent**, which of course **increases** the total volume of the solution.

This means that you can say

#color(blue)( overbrace(c_1 xx V_1)^(color(purple)("moles of solute in initial sol")) = overbrace(c_2 xx V_2)^(color(purple)("moles of solute in target sol")))#

Here

Rearrange to solve for

#c_1 xx V_1 = c_2 xx V_2 implies V_2 = c_1/c_2 * V_1#

In your case, you have

#V_2 = (0.3162 color(red)(cancel(color(black)("M"))))/(0.1995color(red)(cancel(color(black)("M")))) * "25.0 cm"^3 = "39.62 cm"^3#

This means that you must add

#V_2 = V_1 + V_"water"#

#V_"water" = V_2 - V_1 = "39.62 cm"^3 - "25.0 cm"^3 = "14.62 cm"^3#

I'll leave the answer rounded to two sig figs

#V_"water" = color(green)("15 cm"^3)#