Question #9329f

1 Answer
Jan 14, 2017

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.