Question

Question #bf279

Answer 1

`"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.