# Question #bf279

##### 1 Answer

#### Explanation:

**!! VERY LONG ANSWER !!**

Your ultimate goal here is to figure out how much weak acid,

You already know that both buffers have

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

so your starting point here will be to figure out the concentration of

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

#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

#ul("For buffer X")#

This buffer has

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

Rearrange to solve for

#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))#

#ul("For buffer Y")#

This buffer has

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

Rearrange to solve for

#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 **doubles**, the concentration of the weak acid remain **unchanged**.

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 **volume** of the two buffers, and using

#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

#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 remainedunchanged, 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" = 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.