Question

Question #fdeda

Answer 1

Here's how you can do that.

Explanation:

Let's assume that you're dealing with a monoprotic strong acid like hydrochloric acid, `"HCl"`.

Strong acids ionize completely in aqueous solution to form hydronium cations, `"H"_3"O"^(+)`, in a `1:1` mole ratio

`color(red)("H")"Cl"_ ((aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(color(red)(+)) + "Cl"_((aq))^(-)`

This means that every mole of strong acid will produce one mole of hydronium cations in solution. Therefore, for a strong monoprotic acid solution, you have

`["H"_3"O"^(+)] = ["concentration of the acid"]`

In your case, you have

`["HCl"] = "0.0001 mol L"^(-1)`

This means that the solution will contain

`["H"_3"O"^(+)] = "0.0001 mol L"^(-1)`

Since the pH of a solution is defined as the negative log base 10 of the concentration of hydronium cations

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

you can say that the pH of this solution will be

`"pH" = - log(0.0001) = - log(10^(-4)) = 4 + log(1) = color(green)(|bar(ul(color(white)(a/a)4color(white)(a/a)|)))`

Now, an interesting example would be sulfuric acid, `"H"_2"SO"_4`, which is considered, for almost all intended purposes, a diprotic strong acid.

When sulfuric acid ionizes in water, it releases two protons, which means that it produces hydronium cations in a `1:2` mole ratio.

`color(red)("H")_ color(green)(2)"SO"_ (4(aq)) + 2"H"_ 2"O"_ ((l)) -> color(green)(2)"H"_ 3"O"_ ((aq))^(color(red)(+)) + "SO"_(4(aq)^(-)`

This time, you have

`["H"_3"O"^(+)] = 2 xx ["concentration of the acid"]`

In your case, you will end up with

`["H"_3"O"^(+)] = 2 xx "0.0001 mol L"^(-1) = "0.0002 mol L"^(-1)`

The pH will be

`"pH" = - log(0.0002) = - log(2 * 10^(-4)) = 4 + log(2) = color(green)(|bar(ul(color(white)(a/a)3.7color(white)(a/a)|)))`

The pH of this solution is lower than the pH of the first solution because you have a higher concentration of hydronium cations.