Question #75541

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Question #75541

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Answer 1 · 2017-04-13T00:56:05

Here's how you can do that.

Explanation:

Let's say that we're working with a generic monoprotic weak acid $HA$ $"HA"$ .

When you titrate a weak acid with a strong base, which we will represent as ${OH}^{-}$ $"OH"^(-)$ , the following reaction takes place

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

Notice that the acid and the hydroxide anions react in a $1 : 1$ $1:1$ mole ratio. This implies that every mole of hydroxide anions you add to the weak acid solution will consume $1$ $1$ mole of weak acid.

Similarly, $A^{-}$ $"A"^(-)$ , the conjugate base of the weak acid, is produced in a $1 : 1$ $1:1$ mole ratio. This tells you that every mole of hydroxide anions added to the weak acid solution consumes $1$ $1$ mole of $HA$ $"HA"$ and produces $1$ $1$ mole of $A^{-}$ $"A"^(-)$ .

At $50 %$ $50%$ neutralization, half of the moles of $HA$ $"HA"$ have reacted. If you take $n$ $n$ to be the initial number of moles of $HA$ $"HA"$ , then you know that you must add

$n_{{OH}^{-}}^{-} = (\frac{1}{2} \cdot n)$ $n_ ("OH"^(-)) = (1/2 * n)$ ${moles OH}^{-}$ $"moles OH"^(-)$

In order to consume $\frac{1}{2}$ $1/2$ of the number of moles of weak acid you must add $(\frac{1}{2} \cdot n)$ $(1/2 * n)$ moles of hydroxide anions

So, you know that you start with $n$ $n$ moles of $HA$ $"HA"$ . After the reaction is complete, you will have

$n_{HA} = n - (\frac{1}{2} \cdot n) = (\frac{1}{2} \cdot n)$ $n_ ("HA") = n - (1/2 * n) = (1/2 * n)$ $moles HA$ $"moles HA"$

The reaction will produce

$n_{A^{-}}^{-} = 0 + (\frac{1}{2} \cdot n) = (\frac{1}{2} \cdot n)$ $n_ ("A"^(-)) = 0 + (1/2 * n) = (1/2 * n)$ ${moles A}^{-}$ $"moles A"^(-)$

This means that at $50 %$ $50%$ neutralization, i.e. at half equivalence point, the solution will contain equal numbers of moles of weak acid and conjugate base.

This is equivalent to saying that the solution will contain equal concentrations of weak acid and of conjugate base.

$[HA] = [A^{-}] \to$ $["HA" ] =["A"^(-)] ->$ at half equivalence point

You are now in the buffer region of the titration curve, i.e. you have created a weak acid/conjugate base buffer. The Henderson - Hasselbalch equation looks like this

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

In your case, you have

$pH = p K_{a} + log (\frac{[A^{-}]}{[HA]})$ $"pH" = "p"K_a + log( (["A"^(-)])/(["HA"]))$

But since

$[HA] = [A^{-}]$ $["HA" ] =["A"^(-)]$

you can say that

$log( (color(red)(cancel(color(black)(["A"^(-)]))))/(color(red)(cancel(color(black)(["HA"]))))) = log(1) = 0$

This means that the Henderson - Hasselbalch equation becomes

$"pH" = "p"K_a + 0$

and so

$color(darkgreen)(ul(color(black)("pH" = "p"K_a))) ->$ at half equivalence point

The process is exactly the same for a weak base/strong acid titration.

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Question #75541

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