Question #fcf5e
1 Answer
The
Explanation:
For starters, you know that the
Now, the
#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#
This implies that you have
#["H"_3"O"^(+)] = 10^(-"pH")#
Similarly, the
#color(blue)(ul(color(black)("pOH" = 14 - "pH")))#
can be used to find the concentration of hydroxide anions,
#["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
#"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" = 14 - [- log(["OH"^(-)])]#
#"pH" = 14 + log(8.182 * 10^(-2))#
#color(darkgreen)(ul(color(black)("pH" = 12.9)))#
As predicted, the resulting solution has