Question

Question #9329f

Answer 1

Here's why that is the case.

Explanation:

The key here is the definition of a solution's `"pH"`, which as you know is aid to be equal to the negative log base `10` of the concentration of hydronium cations, `"H"_3"O"^(+)`.

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

Now, in order to see why hydrochloric acid is `10` times stronger than citric acid, you need to look at the relative `"pH"` values of the two acids.

You know that you have a difference of `1` unit between the `"pH"` of hydrochloric acid and the `"pH"` of citric acid

`"pH"_ "HCl" = "pH"_ "citric acid" + 1" " " "color(orange)("(*)")`

Use the `"pH"` equation to write

`log(["H"_3"O"^(+)]) = - "pH"`

`10^log(["H"_3"O"^(+)]) = 10^(-"pH")`

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

You can thus say that the concentration of hydronium cations in a solution of hydrochloric acid will be

`["H"_ 3"O"^(+)] _ "HCl" = 10^(-"pH"_ "HCl")`

Use equation `color(orange)("(*)")` to write

`["H"_ 3"O"^(+)]_ "HCl" = 10^(-"pH"_ "citric acid" + 1)`

This is equivalent to

`["H"_ 3"O"^(+)]_ "HCl" = 10^(-"pH"_ "citric acid") * 10^1`

But since

`["H"_ 3"O"^(+)]_ "citric acid" = 10^(-"pH"_ "citric acid")`

it follows that you will have

`color(darkgreen)(ul(color(black)(["H"_ 3"O"^(+)]_ "HCl" = 10 * ["H"_ 3"O"^(+)]_ "citric acid")))`

This shows that a solution of hydrochloric acid is `10` times stronger than an equimolar solution of citric acid because it contains `10` times more hydronium cations.

This is, of course, equivalent to saying that the `"pH"` of the hydrochloric acid solution if `1` unit lower than the `"pH"` of the citric acid solution.