Question #545aa

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Question #545aa

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Answer 1 · 2016-02-26T00:23:18

Here's what I got.

Explanation:

The idea here is that neutralizing a weak acid with a strong base can produce a buffer solution, depending on whether or not the neutralization is complete.

A weak acid will react with a strong base to produce water and the conjugate base of the acid.

If the neutralization is complete, the resulting solution will contain the conjugate base of the acid, which will then react with water to reform some of the weak acid and produce hydronium ions, ${OH}^{-}$ $"OH"^(-)$ , in the process.

If the neutralization is not complete, then the resulting solution will contain a weak acid and its conjugate base $\to$ $->$ a buffer solution.

The pH of a buffer solution that contains a weak acid and its conjugate base can be calculated using the Henderson - Hasselbalch equation

$pH = p K_{a} + log (\frac{[conjugate base]}{[weak acid]})$ $color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))$

Here $p K_{a}$ $pK_a$ represents

$p K_{a} = - log (K_{a})$ $color(blue)(pK_a = - log(K_a))$

So, your problem didn't provide you with concentrations of weak acid and strong base because you don't actually need to know them in order to find the pH of the resulting solution given that a certain percent of the weak acid is neutralized.

Let's assume that your weak acid is $HA$ $"HA"$ and your strong base is $BOH$ $"BOH"$ . A neutralization reaction between the two will look like this

${HA}_{(aq]} + {OH}_{(aq]}^{-} \to A_{(aq]}^{-} + H_{2} O_{(l]}$ $"HA"_text((aq]) + "OH"_text((aq])^(-) -> "A"_text((aq])^(-) + "H"_2"O"_text((l])$

It's important to notice that you have $1 : 1$ $1:1$ mole ratios between the weak acid, the strong base, and the conjugate base $A^{-}$ $"A"^(-)$ .

This means that for every one mole of weak acid neutralized by one mole of strong base, one mole of conjugate base is produced.

Now, let's assume that you want to neutralize $40 %$ $40%$ of an initial concentration of weak acid ${[HA]}_{0}$ $["HA"]_0$ . This is equivalent to saying that $60 %$ $60%$ of the acid will remain present in solution after the reaction is finished.

If you start with a volume $V_{1}$ $V_1$ of weak acid, you will have

$c = \frac{n}{V} \Rightarrow n = c \cdot V$ $color(blue)(c = n/V implies n= c * V)$

$n_{before} = {[HA]}_{0} \cdot V_{1}$ $n_"before" = ["HA"]_0 * V_1$

After the reaction is complete, you need to have

$n_{after} = \frac{6}{10} \cdot {[HA]}_{0} \cdot V_{1} \to$ $n_"after" = 6/10 * ["HA"]_0 * V_1 ->$ moles of weak acid remaining in solution

In order to neutralize $40 %$ $40%$ of the number of moles of acid, you must supply

$n_{base} = \frac{4}{10} \cdot {[HA]}_{0} \cdot V_{1} \to$ $n_"base" = 4/10 * ["HA"]_0 * V_1 ->$ moles of strong base needed for the reaction

Now, if $\frac{4}{10} \cdot {[HA]}_{0} \cdot V_{1}$ $4/10 * ["HA"]_0 * V_1$ moles of weak acid react with $\frac{4}{10} \cdot {[HA]}_{0} \cdot V_{1}$ $4/10 * ["HA"]_0 * V_1$ moles of strong base, the reaction will produce $\frac{4}{10} \cdot {[HA]}_{0} \cdot V_{1}$ $4/10 * ["HA"]_0 * V_1$ moles of $A^{-}$ $"A"^(-)$ $\to$ $->$ remember the $1 : 1$ $1:1$ mole ratios.

Let's say that the volume of strong base is $V_{2}$ $V_2$ . The total volume of the solution will be

$V_{total} = V_{1} + V_{2}$ $V_"total" = V_1 + V_2$

The concentration of the weak acid will now be

$["HA"] = overbrace(6/10 * ["HA"]_0 * V_1)^(color(purple)("moles of weak acid remaining in solution")) * 1/(V_1 + V_2)$

The concentration of the conjugate base will be

$["A"^(-)] = overbrace(4/10 * ["HA"]_0 * V_1)^(color(green)("moles of conjugate base produced")) * 1/(V_1 + V_2)$

Now, plug these values into the Henderson - Hasselbalch equation to get

$"pH" = pK_a + log((["A"^(-)])/(["HA"]))$

$"pH" = pK_a + log( (4/10 * color(red)(cancel(color(black)(["HA"]_0 * V_1))))/(6/10 * color(red)(cancel(color(black)(["HA"]_0 * V_1)))) * color(red)(cancel(color(black)(V_1 + V_2)))/color(red)(cancel(color(black)(V_1 + V_2))))$

$"pH" = pK_a + log(4/color(red)(cancel(color(black)(10))) * color(red)(cancel(color(black)(10)))/6)$

$"pH" = pK_a + log(2/3)$

Now, does this result make sense?

If you neutralize $40%$ of the acid, the resulting solution will contain more weak acid than conjugate base. This means that the pH of the solution will be lower than the $pK_a$ of the weak acid.

Since

$"pH" = pK_a -0.18$

you have will indeed have

$"pH" < pK_a$

As practice, try redoing the calculations if you $75%$ of the acid must be neutralized. This time the pH of the resulting solution must come out to be higher than $pK_a$ .

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Question #545aa

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