Question

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`)?

Answer 1

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.

`color(white)(mmmmml)"HA" color(white)(l)+ color(white)(m)"OH"^"-" → color(white)(ml)"A"^"-" + "H"_2"O"`
`"I/mol:"color(white)(mll)0.025color(white)(mm)0.0080color(white)(mm)0`
`"C/mol:"color(white)(m)"-0.0080"color(white)(m)"-0.0080"color(white)(m)"+0.0080"`
`"E/mol:"color(white)(ml)0.017color(white)(mm)0color(white)(mmmmll)0.0080`

`"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"`

`"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 `"HA"` and 0.0080 mol of `"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.

∴ `"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`