How do I calculate the $pH$ $"pH"$ of the buffer solution formed when $10.0$ $10.0$ ${cm}^{3}$ $"cm"^3$ of $0.80$ $0.80$ $mol$ $"mol"$ ${dm}^{- 3}$ $"dm"^-3$ sodium hydroxide is mixed with $50.0$ $50.0$ ${cm}^{3}$ $"cm"^3$ of $0.50$ $0.50$ $mol$ $"mol"$ ${dm}^{- 3}$ $"dm"^-3$ ethanoic acid ( $K_{a} = 1.74 \times 10^{- 5}$ $"K"_"a"=1.74xx10^-5$ $mol$ $"mol"$ ${dm}^{- 3}$ $"dm"^-3$ )?

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How do I calculate the $pH$ $"pH"$ of the buffer solution formed when $10.0$ $10.0$ ${cm}^{3}$ $"cm"^3$ of $0.80$ $0.80$ $mol$ $"mol"$ ${dm}^{- 3}$ $"dm"^-3$ sodium hydroxide is mixed with $50.0$ $50.0$ ${cm}^{3}$ $"cm"^3$ of $0.50$ $0.50$ $mol$ $"mol"$ ${dm}^{- 3}$ $"dm"^-3$ ethanoic acid ( $K_{a} = 1.74 \times 10^{- 5}$ $"K"_"a"=1.74xx10^-5$ $mol$ $"mol"$ ${dm}^{- 3}$ $"dm"^-3$ )?

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Answer 1 · 2017-04-19T20:07:00

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.

$m m m m m l HA l + m {OH}^{-} \to m l A^{-} + H_{2} O$ $color(white)(mmmmml)"HA" color(white)(l)+ color(white)(m)"OH"^"-" → color(white)(ml)"A"^"-" + "H"_2"O"$
$I/mol: m l l 0.025 m m 0.0080 m m 0$ $"I/mol:"color(white)(mll)0.025color(white)(mm)0.0080color(white)(mm)0$
$C/mol: m -0.0080 m -0.0080 m +0.0080$ $"C/mol:"color(white)(m)"-0.0080"color(white)(m)"-0.0080"color(white)(m)"+0.0080"$
$E/mol: m l 0.017 m m 0 m m m m l l 0.0080$ $"E/mol:"color(white)(ml)0.017color(white)(mm)0color(white)(mmmmll)0.0080$

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

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

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.

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

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