Question

You require 0.100 L of a buffer with pH = 4.35. On the shelf is a bottle of benzoic acid/sodium benzoate buffer with pH = 3.95. ??

The label also says [C7H5O2-] = 0.200 M. What mass of which substance (benzoic acid or sodium benzoate) should you add to 0.100 L of the buffer to obtain the desired pH? Ka of HC7H5O2 = 6.5 x 10-5 .

How would I solve this? I am running short on time...

Answer 1

You must add 4.4 g of sodium benzoate.

Explanation:

You have a buffer with pH = 3.95.

You need a buffer with pH = 4.35 ( more basic).

Thus, you must add a base ( sodium benzoate) to the buffer.

Step 1. Calculate the amounts of benzoic acid and sodium benzoate in 100 mL of the original buffer

The equilibrium involved is

`"HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"`

We can apply the Henderson-Hasselbalch equation:

`color(blue)(bar(ul(|color(white)(a/a)"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))color(white)(a/a)|)))" "`

For benzoic acid,

`K_text(a) = 6.8 × 10^"-5"`

`"p"K_text(a) = "-log"(6.8 × 10^"-5") = 4.17`

`3.95 = 4.17 + log((["A"^"-"])/(["HA"]))`

`log((["A"^"-"])/(["HA"])) = "-0.217"`

`(["A"^"-"])/(["HA"]) = 10^("-0.217") = 0.606`

`["A"^"-"] = 0.606["HA"]`

`["HA"] = (["A"^"-"])/0.606 = "0.200 mol/L"/0.606 = "0.330 mol/L"`

`"Moles of HA" = 0.100 color(red)(cancel(color(black)("L buffer"))) × "0.330 mol HA"/(1 color(red)(cancel(color(black)("L buffer")))) = "0.0330 mol HA"`

`"Moles of A"^"-" = 0.100 color(red)(cancel(color(black)("L buffer"))) × "0.200 mol A"^"-"/(1 color(red)(cancel(color(black)("L buffer")))) = "0.0200 mol A"^"-"`

The original buffer contains 0.0330 mol of benzoic acid and 0.0200 mol of sodium benzoate.

Step 2. Calculate the additional moles of sodium needed for pH 4.35

`4.35 = 4.17 + log((["A"^"-"])/(["HA"]))`

`log((["A"^"-"])/(["HA"])) = "0.183"`

`(["A"^"-"])/(["HA"]) = 10^0.183 =1.52`

`["A"^"-"] = 1.52["HA"] = "1.52 × 0.330 mol/L = 0.502 mol/L"`

`"Moles of A"^"-" = 0.100 color(red)(cancel(color(black)("L buffer"))) × "0.502 mol A"^"-"/(1 color(red)(cancel(color(black)("L buffer")))) = "0.0502 mol A"^"-"`

`"Moles of added sodium benzoate = (0.0502 - 0.0200) mol = 0.0302 mol"`

Step 3. Calculate the mass of the additional sodium benzoate

`"Mass" = 0.0302 color(red)(cancel(color(black)("mol"))) × "144.11 g"/(1 color(red)(cancel(color(black)("mol")))) = "4.4 g"`