A solution is made by mixing 500mL of .3M benzoic acid and 400mL of .25M benzoate. What will the pH of the solution be?. Specifically, how do I find the concentration of the acid and base to use in the equation #pH=pKa +log( ([A^(-)])/([HA]))# ?

#K_a = 6.46 xx 10^(-5)#

2 Answers

Answer:

#"pH" = 4.0#

Explanation:

As you know, a solution's molarity, or molar concentration, is calculated by taking the number of moles of solute present in #"1 L"# of solution.

This basically means that in order to find a solution's molarity, you must know

  • how many moles of solute it contains
  • the total volume of the solution

The first thing you need to do here is to calculate how many moles of benzoic acid, #"C"_7"H"_6"O"_2#, a weak acid, and of benzoate anions, #"C"_7"H"_5"O"_2^(-)#, its conjugate base, you have in the two solutions that you're mixing.

You will have

#500 color(red)(cancel(color(black)("mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.3 moles C"_7"H"_6"O"_2)/(1color(red)(cancel(color(black)("L solution")))#

#= "0.150 moles C"_7"H"_6"O"_2#

and

#400 color(red)(cancel(color(black)("mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.25 moles C"_7"H"_5"O"_2^(-))/(1color(red)(cancel(color(black)("L solution")))#

#="0.100 moles C"_7"H"_5"O"_2^(-)#

All you need now is the total volume of the resulting solution, which you can get by adding the volumes of the two solutions

#V_"total" = "500 mL" + "400 mL" = "900 mL"#

The concentrations of the two species in the resulting solution will be -- remember to use the volume in liters !

#["C"_7"H"_6"O"_2] = "0.150 moles"/(900 * 10^(-3)"L") = "0.1667 M"#

#["C"_7"H"_5"O"_2^(-)] = "0.100 moles"/(900 * 10^(-3)"L") = "0.1111 M"#

Now you're ready to use the Henderson - Hasselbalch equation

#"pH" = "p"K_a + log( (["C"_7"H"_5"O"_2^(-)])/(["C"_7"H"_6"O"_2]))#

Plug in your values to find

#"pH" = -log(6.46 * 10^(-5)) + log( (0.1111 color(red)(cancel(color(black)("M"))))/(0.1667color(red)(cancel(color(black)("M")))))#

#color(darkgreen)(ul(color(black)("pH" = 4.0)))#

Notice that

#"p"K_a = - log(6.46 * 10^(-5)) = 4.2#

and

#"pH" < "p"K_a#

This is the case because the concentration of the weak acid is higher than the concentration of the conjugate base in the resulting solution

#["C"_7"H"_6"O"_2] > ["C"_7"H"_5"O"_2^(-)] implies "pH" < "p"K_a#

Dec 30, 2016

You know that the main process is mixing these two solutions together. Doing that will dilute them (benzoic acid, #"HA"#, and, say, sodium benzoate, #"A"^(-)#) to a new total volume (imagine doing this in real life).

It is reasonable to assume that their total volume is additive, so that it is #"500 + 400 = 900 mL"#. To find their new concentrations, note that the #"mol"#s of each compound will not change, but their concentrations do.

So, find the #"mol"#s of each compound in the original two solutions. The #"mol"#s of each component are found from its initial molar concentration #[" "]_0#:

#["Component"]_0 = n_("Component")/(V_"Component")#

#=> n_"Component" = ["Component"]_0 cdotV_"Component"#

Therefore:

#n_("HA") = ["HA"]_0 cdotV_("HA") = ("0.300 M")(500xx10^(-3) "L") = "0.150 mols HA"#

#n_("A"^(-)) = ["A"^(-)]_0 cdotV_("A"^(-)) = ("0.250 M")(400xx10^(-3) "L") = "0.100 mols A"^(-)#

Then, use these well-defined quantities (as in, they don't change because there is no meaningful reaction) to find the components' new concentrations #[" "]# in the combined solution:

#color(green)(["HA"]) = "0.150 mols HA"/(("500 + 400") xx 10^(-3) "L") = color(green)("0.166 M")#

#color(green)(["A"^(-)]) = "0.100 mols HA"/(("500 + 400") xx 10^(-3) "L") = color(green)("0.111 M")#

This allows you to find their #"pH"# via the Henderson-Hasselbalch equation:

#color(blue)("pH") = "pKa" + log((["A"^(-)])/(["HA"]))#

#= -log(K_a) + log((["A"^(-)])/(["HA"]))#

#= -log(6.46xx10^(-5)) + log("0.111 M"/"0.166 M")#

#= color(blue)(4.01)#

Or, if you recall that these compounds are both in the same solution, you may recognize that you can also use the #"mol"#s, since the total volume cancels out:

#color(blue)("pH") = -log(6.46xx10^(-5)) + log("0.100 mols"/"0.150 mols")#

#= color(blue)(4.01)#

Either way, we get the same result: #"pH" < "pKa"#, which tells you that the solution is particularly acidic.

Therefore, there is more of the acid form (#"HA"#) than the conjugate base (#"A"^(-)#) in the final solution. You can check that by examining the concentrations in the logarithm.


Depending on how confident you feel about finding concentrations, you may want to pick one method over the other. It can be easy to forget that these solutions have a new total volume.