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

## a solution that contains 1.40% ${C}_{2} {H}_{5} N {H}_{2}$ by mass and 1.18% ${C}_{2} {H}_{5} N {H}_{3} B r$ by mass a solution that is 12.5 g of $H {C}_{2} {H}_{3} {O}_{2}$ and 14.0 g of $N a {C}_{2} {H}_{3} {O}_{2}$ in 150.0 mL of solution?

May 11, 2016

#### Explanation:

1. The basic buffer

The chemical equilibrium for a basic buffer is

$\text{B + H"_2"O" ⇌ "BH"^+ + "OH"^"-}$

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, ($\text{B}$), $\text{p"K_"b} = 3.19$.

Since both $\text{B}$ and ${\text{BH}}^{+}$ 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"^+).

$\text{Moles of B" = 14.0 color(red)(cancel(color(black)("g B"))) × "1 mol B"/(45.08 color(red)(cancel(color(black)("g B")))) = "0.3106 mol B}$

${\text{Moles BH"^+ = 11.8 color(red)(cancel(color(black)("g BH"^+))) × ("1 mol BH"^+)/(126.00 color(red)(cancel(color(black)("g BH"^+)))) = "0.093 65 mol BH}}^{+}$

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

$\text{pH" = "14.00 – pOH" = "14.00 – 2.67} = 11.33$

2. The acidic buffer

The chemical equilibrium for an acidic buffer is

$\text{HA" + "H"_2"O" ⇌ "H"_3"O"^+ +"A"^"-}$

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 ($\text{HA}$), $\text{p"K_"a} = 4.75$.

Since both $\text{HA}$ and $\text{A"^"-}$ 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"^"-").

$\text{Moles of HA" = 12.5 color(red)(cancel(color(black)("g HA"))) × "1 mol HA"/(60.05 color(red)(cancel(color(black)("g HA")))) = "0.2082 mol HA}$

$\text{Moles of A"^"-"=14.0 color(red)(cancel(color(black)("g A"^"-"))) × (1 "mol A"^"-")/(82.03 color(red)(cancel(color(black)("g A"^"-")))) = "0.1707 mol A"^"-}$

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