Question #135b1
1 Answer
Here's what I got.
Explanation:
A weak monoprotic acid labeled
#"HX"_text((aq]) + "H"_2"O"_text((aq]) rightleftharpoons "X"_text((aq])^(-) + "H"_3"O"_text((aq])^(+)#
Your strategy here will be to use the
#color(blue)(pK_a = - log(K_a) implies K_a = 10^(-pK_a))#
This will get you
#K_a = 10^(-3.72) = 1.9 * 10^(-4)#
The ICE table for this equilibrium will look like this
#" ""HX"_text((aq]) + "H"_2"O"_text((aq]) " "rightleftharpoons" " "X"_text((aq])^(-) " "+" " "H"_3"O"_text((aq])^(+)#
You know that
#color(blue)(K_a = (["X"^(-)] * ["H"_3"O"^(+)])/(["HX"]))#
In your case, you will have
#K_a = (x * x)/(0.52 - x) = x^2/(0.52 -x) = 1.9 * 10^(-4)#
Because
#0.52 - x ~~ 0.52#
This will get you
#1.9 * 10^(-4) = x^2/0.52#
Solve for
#x = sqrt(0.52 * 1.9 * 10^(-4)) = 9.94 * 10^(-3)#
Since
#["H"_3"O"^(+)] = x = 9.94 * 10^(-3)#
The pH of the solution will thus be
#color(blue)("pH" = - log(["H"_3"O"^(+)]))#
#"pH" = - log(9.94 * 10^(-3)) = color(green)(2.0)#
Now, what happens when half of the acid is neutralized by a strong base like sodium hydroxide?
The acid will react with the base, which I'll represent as
#"HX"_text((aq]) + "OH"_text((aq])^(-) -> "X"_text((aq])^(-) + "H"_2"O"_text((aq])#
Now, for every mole of strong base added to the acid solution, one mole of weak acid will be consumed and one mole of conjugate base will be produced.
Neutralizing half of the acid is equivalent to converting half of its moles to moles of water and moles of conjugate by adding strong base.
So if
This will of course produce
The resulting solution will contain
#color(blue)(["HX"] = ["X"^(-)]) -># when half of the acid has been neutralized
Plug this into the expression for
#K_a = ( color(red)(cancel(color(black)(["X"^(-)]))) * ["H"_3"O"^(+)])/color(red)(cancel(color(black)(["HX"])))#
#K_a = ["H"_3"O"^(+)]#
Therefore,
#"pH" = - log(["H"_3"O"^(+)]) = overbrace(- log(K_a))^(color(purple)(pK_a))#
#"pH" = color(green)(pK_a)#
And there you have it. When equal concentrations of weak acid and conjugate base are present in solution, the pH of the buffer is equal to the acid's