Question

Question #afe36

Answer 1

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, `"H"_3"O"^(+)`, are present in the initial solution.

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

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 `0.7`

`["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

`c_1`, `V_1` - the concentration and volume of the initial solution
`c_2`, `V_2` - the concentration and volume of the target solution

Rearrange to solve for `V_2`

`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)`