Question #bf279

1 Answer
Dec 6, 2016

Answer:

#"pH" = 5.7#

Explanation:

!! VERY LONG ANSWER !!

Your ultimate goal here is to figure out how much weak acid, #"HA"#, and how much conjugate base, #"A"^(-)#, you have in the target solution.

You already know that both buffers have

#["HA"] = "0.5 M"#

so your starting point here will be to figure out the concentration of #"A"^(-)# in buffer #"X"# and in buffer #"Y"#.

To do that, use the Henderson - Hasselbalch equation, which for a weak acid - conjugate base buffer takes the form

#color(blue)(ul(color(black)("pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))))#

Calculate the #"p"K_a# of the acid by using

#color(blue)(ul(color(black)("p"K_a = - log(K_a))))#

In your case, you will have

#"p"K_a = - log(1.0 * 10^(-5)) = 5.0#

Now, calculate the concentration of #"A"^(-)# in the two buffers

#color(white)(a)#

  • #ul("For buffer X")#

This buffer has #"pH" = 4.0#, which means that you get

#4.0 = 5.0 + log((["A"^(-)]_"X")/(["HA"]))#

Rearrange to solve for #["A"^(-)]_"X"/(["HA"])#

#log(["A"^(-)]_"X"/(["HA"])) = -1.0#

#10^log(["A"^(-)]_"X"/(["HA"])) = 10^(-1.0)#

This will get you

#["A"^(-)]_"X"/(["HA"]) = 0.10#

which results in

#["A"^(-)]_"X" = 0.10 * ["HA"]#

#["A"^(-)]_"X" = 0.10 * "0.5 M" = "0.050 M" " "color(orange)((1))#

#color(white)(a)#

  • #ul("For buffer Y")#

This buffer has #"pH" = 6.0#, which means that you get

#6.0 = 5.0 + log((["A"^(-)]_"Y")/(["HA"]))#

Rearrange to solve for #["A"^(-)]_"Y"/(["HA"])#

#log(["A"^(-)]_"Y"/(["HA"])) = 1.0#

This will get you

#["A"^(-)]_"Y"/(["HA"]) = 10#

which results in

#["A"^(-)]_"Y" = 10 * ["HA"]#

#["A"^(-)]_"Y" = 10 * "0.5 M" = "5.0 M" " "color(orange)((2))#

Now, you are told that you must mix equal volumes of buffer #"X"# and of buffer #"Y"#. Right from the start, you could say that because the volume doubles, the concentration of the weak acid remain unchanged. #color(purple)(("*"))#

That is the case because you're essentially doubling the number of moles of weak acid and the volume of the solution.

If you take #x# #"L"# to be the volume of the two buffers, and using #color(orange)((1))# and #color(orange)((2))#, you can say that buffer #"X"# will contain

#x color(red)(cancel(color(black)("L buffer"))) * "0.5 moles HA"/(1color(red)(cancel(color(black)("L buffer")))) = (0.5 * x)" moles HA"#

#x color(red)(cancel(color(black)("L buffer"))) * "0.050 moles A"^(-)/(1color(red)(cancel(color(black)("L buffer")))) = (0.050 * x)" moles A"^(-)#

Similarly, buffer #"Y"# will contain

#x color(red)(cancel(color(black)("L buffer"))) * "0.5 moles HA"/(1color(red)(cancel(color(black)("L buffer")))) = (0.5 * x)" moles HA"#

#x color(red)(cancel(color(black)("L buffer"))) * "5.0 moles A"^(-)/(1color(red)(cancel(color(black)("L buffer")))) = (5.0 * x)" moles A"^(-1)#

The resulting solution, which has a volume of

#xcolor(white)(.)"L" + xcolor(white)(.)"L" = 2xcolor(white)(.)"L"#

will contain

#{((0.5x)" moles" + (0.5x)" moles" = x" moles HA"), ((0.050x)" moles" + (5.0x)" moles" = 5.05x" moles A"^(-)) :}#

The concentrations of the two species in the resulting solution will be

#["HA"] = (color(red)(cancel(color(black)(x)))"moles")/(2color(red)(cancel(color(black)(x)))"L") = "0.5 M" -># the concentration remained unchanged, as predicted in #color(purple)(("*"))#

#["A"^(-)] = (5.05color(red)(cancel(color(black)(x)))"moles")/(2color(red)(cancel(color(black)(x)))"L") = "2.525 M"#

Finally, use the Henderson - Hasselbalch equation to find the #"pH"# of the resulting solution

#"pH" = 5.0 + log( (2.525 color(red)(cancel(color(black)("M"))))/(0.5color(red)(cancel(color(black)("M"))))) = color(darkgreen)(ul(color(black)(5.7)))#

The answer is rounded to one decimal place, since that is how many sig figs you have for the concentration of the weak acid.