Question

How would you use the Henderson–Hasselbalch equation to calculate the pH of each solution?

1. a solution that contains 1.40% `C_2H_5NH_2` by mass and 1.18% `C_2H_5NH_3Br` by mass

2. a solution that is 12.5 g of `HC_2H_3O_2` and 14.0 g of `NaC_2H_3O_2` in 150.0 mL of solution?

Answer 1

Warning! Long answer.

Explanation:

1. The basic buffer

The chemical equilibrium for a basic buffer is

`"B + H"_2"O" ⇌ "BH"^+ + "OH"^"-"`

The Henderson-Hasselbalch equation for this buffer is

`color(blue)(|bar(ul(color(white)(a/a) "pOH" = "p"K_"b" + log((["BH"^+])/(["B"]))color(white)(a/a)|)))" "`

For ethylamine, (`"B"`), `"p"K_"b" = 3.19`.

Since both `"B"` and `"BH"^+` are in the same solution, the ratio of their molarities is the same as the ratio of their moles.

Assume that we have 1000 g of solution.

Then we have 14.0 g `"CH"_3"CH"_2"NH"_2 ("B")` and 11.8 g `"CH"_3"CH"_2"NH"_3"Br" ("BH"^+)`.

`"Moles of B" = 14.0 color(red)(cancel(color(black)("g B"))) × "1 mol B"/(45.08 color(red)(cancel(color(black)("g B")))) = "0.3106 mol B"`

`"Moles BH"^+ = 11.8 color(red)(cancel(color(black)("g BH"^+))) × ("1 mol BH"^+)/(126.00 color(red)(cancel(color(black)("g BH"^+)))) = "0.093 65 mol BH"^+`

`"pOH" = "p"K_b + log((["BH"^+])/(["B"])) = 3.19 + log(("0.093 65" color(red)(cancel(color(black)("mol"))))/(0.3106 color(red)(cancel(color(black)("mol")))))=3.19 + log(0.3015) = "3.19 – 0.52" = 2.67`

`"pH" = "14.00 – pOH" = "14.00 – 2.67" = 11.33`

2. The acidic buffer

The chemical equilibrium for an acidic buffer is

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

The Henderson-Hasselbalch equation for this buffer is

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

For acetic acid (`"HA"`), `"p"K_"a" = 4.75`.

Since both `"HA"` and `"A"^"-"` are in the same solution, the ratio of their molarities is the same as the ratio of their moles.

We have 12.5 g `"HC"_2"H"_3"O"_2 ("HA")` and 14.0 g `"NaC"_2"H"_3"O"_2 ("A"^"-")`.

`"Moles of HA" = 12.5 color(red)(cancel(color(black)("g HA"))) × "1 mol HA"/(60.05 color(red)(cancel(color(black)("g HA")))) = "0.2082 mol HA"`

`"Moles of A"^"-"=14.0 color(red)(cancel(color(black)("g A"^"-"))) × (1 "mol A"^"-")/(82.03 color(red)(cancel(color(black)("g A"^"-")))) = "0.1707 mol A"^"-"`

`"pH" = "p"K_"a" + log((["A"^"-"])/(["HA"])) = 4.75 + log((0.1707 color(red)(cancel(color(black)("mol"))))/(0.2082 color(red)(cancel(color(black)("mol"))))) = 4.75 + log(0.8199) = "4.75 – 0.09" = 4.66`