Question

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)`

Answer 1

`"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`

Answer 2

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.

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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.