Question #9988f
1 Answer
Here's what I got.
Explanation:
!! VERY LONG ANSWER !!
The idea here is that the nitrite anion,
#"pH" > 7" " and " " "pOH" < 7#
The nitrite anions are delivered to the solution by the soluble sodium nitrite in a
#["NO"_ 2^(-)]_0 = "0.125 M"#
Now, when the nitrite anion is present in aqueous solution, the following equilibrium is established
#"NO"_ (2(aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HNO"_ (2(aq)) + "OH"_ ((aq))^(-)#
The base dissociation constant,
#color(blue)(ul(color(black)(K_a * K_b = 1 * 10^(-14))))#
In your case, the acid dissociation of the nitrous acid,
#K_b = (1 * 10^(-14))/(6.5 * 10^(-4)) = 1.54 * 10^(-11)#
Now, by definition, the base dissociation constant is equal to
#K_b = (["HNO"_2] * ["OH"^(-)])/(["NO"_2^(-)])#
Notice that the nitrite anions react in
#["NO"_ 2^(-)] = ["NO"_ 2^(-)]_ 0 - x# This basically means that in order for the reaction to produce
#x# #"M"# of nitrous acid and hydroxide anions, it must consume#x# #"M"# of nitrite anions.
This means that you have
#K_b = (x * x)/(0.125 - x)#
Since the value of the base dissociation constant is so small compared with the initial concentration of the nitrite anions, which means that the equilibrium lies far to the left, you can use the approximation
#0.125 - x ~~ 0.125#
You will thus have
#K_b = x^2/0.125#
which will get you
#x = sqrt( (0.125 * 1.54 * 10^(-11))#
#x = 1.39 * 10^(-6)#
Since
#["OH"^(-)] = color(darkgreen)(ul(color(black)(1.39 * 10^(-6)color(white)(.)"M")))#
Using the fact that an aqueous solution at
#color(blue)(ul(color(black)(["H"_3"O"^(+)] * ["OH"^(-)] = 1 * 10^(-14)color(white)(.)"M"^2)))#
you can say that this solution will have
#["H"_3"O"^(+)] = (1 * 10^(-14) "M"^color(red)(cancel(color(black)(2))))/(1.39 * 10^(-6)color(red)(cancel(color(black)("M")))) = color(darkgreen)(ul(color(black)(7.19 * 10^(-9)color(white)(.)"M")))#
Consequently, you will have
#"pOH" = - log(["OH"^(-)])#
#"pOH" = - log(1.39 * 10^(-6)) = color(darkgreen)(ul(color(black)(5.86)))#
and
#"pH" = - log (["H"_3"O"^(+)])#
#"pH" = - log(7.19 * 10^(-9))#
#"pH" = color(darkgreen)(ul(color(black)(8.14)))#
Notice that you get the same result by using
#color(blue)(ul(color(black)("pH + pOH" = 14)))#
The equilibrium concentrations are rounded to three sig figs, but I'll leave the
As predicted, the
A MORE ACCURATE APPROACH
An interesting thing to keep in mind here is that pure water already contains hydronium and hydroxide anions
At
#color(blue)(ul(color(black)(["H"_3"O"^(+)] = ["OH"^(-)] = 1 * 10^(-7)color(white)(.)"M")))#
Now, notice that the equilibrium concentration of hydroxide anions that we calculated
#["OH"^(-)] = 1.39 * 10^(-6)color(white)(.)"M"#
is actually quite small and comparable to the initial concentration of hydroxide anions, i.e.
To account for this, you can use the fact that the solution already contains hydroxide anions when you dissolve the salt. So if you use
#1 * 10^(-7) + x# This means that the concentration of hydroxide anions starts at
#1 * 10^(-7)# in pure water and increases by#x# #"M"# when#x# #"M"# of nitrite anions react.
The base dissociation constant will now be
#K_b = (x * (1 * 10^(-7) + x))/(0.125 - x)#
Using the same approximation as before, you will end up with
#x^2 + 1 * 10^(-7) * x - 0.125 * 1.54 * 10^(-11) = 0#
This quadratic equation will produce two solutions, one positive and one negative. Since we're looking for concentration here, you can discard the negative solution to say that
#x = 1.34 * 10^(-6)color(white)(.)"M"#
This time, you will have
#["OH"^(-)] = 1.34 * 10^(-6)color(white)(.)"M"#
#["H"_3"O"^(+)] = 7.46 * 10^(-9)color(white)(.)"M"#
#"pOH" = 5.87#
#"pH" = 8.13#
Now, because the percent error that you get by disregarding the initial concentration of hydroxide anions is
#"% error" = (1.39 * color(red)(cancel(color(black)(10^(-6)color(white)(.)"M"))) - 1.34 * color(red)(cancel(color(black)(10^(-6)color(white)(.)"M"))))/(1.34 * color(red)(cancel(color(black)(10^(-6)color(white)(.)"M")))) * 100% = 3.73% < color(red)(5%)#
you can use the values that you got without accounting for the initial concentration of the hydroxide anions.