Question

Question #fcf5e

Answer 1

The `"pH"` of the solution decreases.

Explanation:

For starters, you know that the `"pH"` of the solution will decrease because you're adding hydrochloric acid, a strong acid. So you should expect the `"pH"` of the solution to be `< 13`.

Now, the `"pH"` of a solution is given by the concentration of hydronium cations, `"H"_3"O"^(+)`.

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

This implies that you have

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

Similarly, the `"pOH"` of a solution, which for an aqueous solution at room temperature is equal to

`color(blue)(ul(color(black)("pOH" = 14 - "pH")))`

can be used to find the concentration of hydroxide anions, `"OH"^(-)`.

`["OH"^(-)] = 10^(-"pOH")`

This means that the sodium hydroxide solution has

`["OH"^(-)] = 10^(14 - 13)` `"M"`

`["OH"^(-)] = 10^(-1)` `"M"`

The number of moles of hydroxide anions present in your sample is equal to

`100 color(red)(cancel(color(black)("mL solution"))) * (10^(-1)color(white)(.)"moles OH"^(-))/(10^3color(red)(cancel(color(black)("mL solution")))) = 10^(-2)color(white)(.)"moles OH"^(-)`

The hydrochloric acid solution has

`["H"_3"O"^(+)] = 10^(-1)` `"M"`

which means that the sample contains

`10 color(red)(cancel(color(black)("mL solution"))) * (10^(-1)color(white)(.)"moles H"_3"O"^(+))/(10^3color(red)(cancel(color(black)("mL solution")))) = 10^(-3)color(white)(.)"moles H"_3"O"^(+)`

Sodium hydroxide and hydrochloric acid neutralize each other in a `1:1` mole ratio

`"OH"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+) -> 2"H"_ 2"O"_ ((l))`

so you can say that when you mix these two solutions, the hydronium cations present in the hydrochloric acid solution will be the limiting reagent, i.e. they will be completely consumed by the reaction.

You can thus say that the reaction will leave you with

`10^(-3)color(white)(.)"moles" - 10^(-3)color(white)(.)"moles" = "0 moles H"_3"O"^(+)`

and with

`10^(-2)color(white)(.)"moles" - 10^(-3)"moles" = 9 * 10^(-3)color(white)(.)"moles OH"^(-)`

The total volume of the solution will be

`V_"total" = "100 mL + 10 mL = 110 mL"`

This means that the concentration of the hydroxide anions in the resulting solution will be

`["OH"^(-)] = (9 * 10^(-3)color(white)(.)"moles")/(110 * 10^(-3)color(white)(.)"L") = 8.182 * 10^(-2)` `"M"`

Therefore, the `"pH"` of the resulting solution will be

`"pH" = 14 - [- log(["OH"^(-)])]`

`"pH" = 14 + log(8.182 * 10^(-2))`

`color(darkgreen)(ul(color(black)("pH" = 12.9)))`

As predicted, the resulting solution has `"pH" < 13`, which is consistent with the fact that you're adding a strong acid to the initial solution.