Calculate the #"pH"# of a solution that resulted when #"40 mL"# of a #"0.1-M"# ammonia solution was diluted to #"60 mL"# ?
The #K_a# of the ammonium cation is #5.7 * 10^-10#
The
1 Answer
Explanation:
Start by calculating the molarity of the diluted ammonia solution.
You know that your stock solution contains
#40 color(red)(cancel(color(black)("mL solution"))) * "0.1 moles NH"_3/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0040 moles NH"_3#
After you dilute this solution by adding enough water to increase its volume from
#["NH"_3] = "0.0040 moles"/(60 * 10^(-3) quad "L") = "0.0667 M"#
Now, ammonia will act as a weak base in aqueous solution.
#"NH"_ (3(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "NH"_ (4(aq))^(+) + "OH"_ ((aq))^(-)#
By definition, the expression of the base dissociation constant,
#K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3^(+)])#
As you know, an aqueous solution at
#color(blue)(ul(color(black)(K_a * K_b = 1 * 10^(-14))))#
Notice that the problem gives you the acid dissociation constant,
#K_b= (1 * 10^(-14))/(5.7 * 10^(-10)) = 1.75 * 10^(-5)#
Now, if you take
#["NH"_3] = (0.0667 - x) quad "M"# This basically means that in order for the reaction to produce
#x# #"M"# of ammonium cations and#x# #"M"# of hydroxide anions, the concentration of ammonia must decrease by#x# #"M"# .
Plug this back into the expression of the base dissociation constant to find
#1.75 * 10^(-5) = (x * x)/(0.0667 - x)#
#1.75 * 10^(-5) = x^2/(0.0667 - x)#
The value of the base dissociation constant is small enough compared to the initial concentration of the acid to justify the approximation
#0.0667 - x ~~ 0.0667#
This means that you have
#1.75 * 10^(-5) = x^2/0.0667#
which gets you
#x = sqrt(0.0667 * 1.75 * 10^(-5)) = 1.08 * 10^(-3)#
Since
#["OH"^(-)] = 1.08 * 10^(-3) quad "M"#
Consequently, the
#color(blue)(ul(color(black)("pH + pOH = 14")))#
will be equal to
#"pH" = 14 - "pOH"#
Since
#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))#
you can say that your solution has
#"pH" = 14 - [- log(1.08 * 10^(-3))] = color(darkgreen)(ul(color(black)(11.0)))#
The answer is rounded to one decimal place, the number of sig figs you have for your values.