What is the pH of a solution made of 0.1 M acetic acid and 0.1 M potassium acetate to which 0.001 mol of KOH has been added?

Ka = 1.8 x 10-5

a)pH = 8.92
b)pH = 4.74
c)pH = 11.55
d)pH = 4.46
e)pH = 4.73

I tried subtracting .001 moles from .1 moles from acetic acid and adding .001 moles to potassium acetate (I assumed that there was 1 liter of solution) and when I use the Henderson-Hasselbalch equationI get 4.74. The practice quiz is marking it wrong, so I'm not sure what I'm doing wrong...

1 Answer
Mar 14, 2018

Answer:

Here's what I got.

Explanation:

For starters, I think that the question is either missing the option #4.75# or one of the options given to you was mistyped.

The idea here is that adding a very small amount of strong base to the buffer will increase the #"pH"# of the buffer ever so slightly.

Even without doing any calculations, you can say that adding #0.001# moles of potassium hydroxide would be enough to increase the #"pH"# of the buffer by #0.01#, from #4.74#, what you have at first, to #4.75#.

As you know, the #"pH"# of a weak acid - conjugate base buffer can be calculated using the Henderson - Hasselbalch equation.

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

Now, when you have equal concentrations of the weak acid and of the conjugate base, you get

#log( (["conjugate base"])/(["weak acid"])) = log(1) = 0#

and

#"pH" = "p"K_a#

So before adding the potassium hydroxide, your buffer has

#["CH"_3"COOH"] = ["CH"_3"COO"^(-)] = "0.1 M"#

and

#"pH" = - log(1.8 * 10^(-5))#

#"pH" = 4.74#

Now, adding the strong base will cause the concentration of the weak acid to decrease and the concentration of the conjugate base to increase.

So after you add the potassium hydroxide, you have

#["CH"_3"COOH"] < ["CH"_3"COO"^(-)]#

This means that

#log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) > 0#

and so

#"pH" = 4.74 + ("something > 0")#

which gets you

#"pH" > 4.74#

Since the number of moles of potassium hydroxide is very small compared to the initial concentration of the weak acid and of the conjugate base (assuming a #"1-L"# solution), you can say that an option like #4.75# would be an ideal fit here.

#color(darkgreen)(ul(color(black)("pH" > 4.74 " " -> " but very close to 4.74")))#

You can follow the same logic and say that if you start with the same buffer and add #0.001# moles of hydrochloric acid, a strong acid, the #"pH"# would be #4.73#.

This time, the strong acid would consume some of the conjugate base and produce weak acid, so

#(["CH"_3"COOH"] > ["CH"_3"COO"^(-)]#

This means that

#log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) < 0#

and so

#"pH" = 4.74 + ("something < 0")#

which gets you

#"pH" < 4.74#

In this case, #4.73# would be an ideal fit because the number of moles of hydrochloric acid is very small compared to the initial concentration of the weak acid and of the conjugate base (assuming a #"1-L"# solution).

#color(darkgreen)(ul(color(black)("pH" < 4.74 " "->" but very close to 4.74")))#

Now, back to the problem at hand. If you assume that you're working with a #"1-L"# solution and that the addition of the strong base does not change the volume of the buffer, you can use the molarity and the number of moles interchangeably to say that after you add #0.001# moles of potassium hydroxide, you will have

#{((0.1 - 0.001) quad "moles CH"_3"COOH"), ((0.1 + 0.001) quad "moles CH"_3"COO"^(-)) :}#

And so

#"pH" = "p"K_a + log (((0.1 + 0.001) color(red)(cancel(color(black)("moles"))))/((0.1 - 0.001) color(red)(cancel(color(black)("moles")))))#

#"pH" = 4.7487 ~~ 4.75#

As predicted, we have

#"pH" = 4.75 > 4.74 " " -> " but very close to 4.74"#