# How do I calculate the "pH" of the buffer solution formed when 10.0 "cm"^3 of 0.80 "mol" "dm"^-3 sodium hydroxide is mixed with 50.0 "cm"^3 of 0.50 "mol" "dm"^-3 ethanoic acid ("K"_"a"=1.74xx10^-5 "mol" "dm"^-3)?

Apr 19, 2017

The pH of the buffer is 4.43.

#### Explanation:

The first thing you must recognize is that there will be an acid-base neutralization reaction.

Your first task is to calculate the concentrations of the species present at the end of the reaction.

We can summarize the calculations in an ICE table.

$\textcolor{w h i t e}{m m m m m l} \text{HA" color(white)(l)+ color(white)(m)"OH"^"-" → color(white)(ml)"A"^"-" + "H"_2"O}$
$\text{I/mol:} \textcolor{w h i t e}{m l l} 0.025 \textcolor{w h i t e}{m m} 0.0080 \textcolor{w h i t e}{m m} 0$
$\text{C/mol:"color(white)(m)"-0.0080"color(white)(m)"-0.0080"color(white)(m)"+0.0080}$
$\text{E/mol:} \textcolor{w h i t e}{m l} 0.017 \textcolor{w h i t e}{m m} 0 \textcolor{w h i t e}{m m m m l l} 0.0080$

$\text{Moles HA" = 0.0500 color(red)(cancel(color(black)("dm"^3 color(white)(l)"HA"))) × "0.50 mol HA"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"HA")))) = "0.025 mol HA}$

$\text{Moles OH"^"-" = 0.0100 color(red)(cancel(color(black)("dm"^3 color(white)(l)"A"^"-"))) × "0.80 mol A"^"-"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"A"^"-")))) = "0.0080 mol A"^"-}$

So, at the end of the reaction, we have a solution containing 0.017 mol of $\text{HA}$ and 0.0080 mol of $\text{A"^"-}$.

A mixture of a weak acid and its salt is a buffer.

We can use the Henderson-Hasselbalch equation to calculate the pH of the buffer:

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

Since both components are in the same solution, the ratio of their concentrations is the same as the ratio of their moles.

$\text{pH" = "-log"(1.74 × 10^"-5") + log((0.0080 color(red)(cancel(color(black)("mol"))))/(0.017 color(red)(cancel(color(black)("mol"))))) = "4.76 - 0.33} = 4.43$