Suppose you have $500.0 mL$ $"500.0 mL"$ of $0.200 M$ $"0.200 M"$ acetic acid ( $K_{a} = 1.8 \times 10^{- 5}$ $K_a = 1.8 xx 10^(-5)$ ). To make a $pH$ $"pH"$ $4.50$ $4.50$ buffer, how many $mL$ $"mL"$ of $3.0 M$ $"3.0 M"$ $NaOH$ $"NaOH"$ must be added?

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Suppose you have $500.0 mL$ $"500.0 mL"$ of $0.200 M$ $"0.200 M"$ acetic acid ( $K_{a} = 1.8 \times 10^{- 5}$ $K_a = 1.8 xx 10^(-5)$ ). To make a $pH$ $"pH"$ $4.50$ $4.50$ buffer, how many $mL$ $"mL"$ of $3.0 M$ $"3.0 M"$ $NaOH$ $"NaOH"$ must be added?

This is a fun challenge question this week in the chemistry class I'm TAing. I thought I might as well put it up here and answer it.

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Answer 1 · 2017-03-21T05:42:55

There are two ways to approach this problem:

Construct an ICE table to see what ${[{OAc}^{-}]}_{e q}^{-}$ $["OAc"^(-)]_(eq)$ is after using the small $x$ $x$ approximation, and include ${[{OAc}^{-}]}_{e q}^{-}$ $["OAc"^(-)]_(eq)$ in the calculations. This gives the most accurate answer, unless you wish to find the true $x$ $x$ without the small $x$ $x$ approximation!
Ignore the ICE table altogether and therefore assume that $n_{{OH}^{-}}^{-}$ $n_("OH"^(-))$ will overshadow $n_{{OAc}^{-}}^{-}$ $n_("OAc"^(-))$ in solution. This is the fastest way to the answer, but gives a $2.63 %$ $2.63%$ error in comparison to the first method with the small $x$ $x$ approximation.

I'll show the results of both methods (including the exact quadratic formula answer).

DISCLAIMER: LONG ANSWER!

METHOD 1

The acid placed into water reacts as follows:

$HOAc (a q) + H_{2} O (l) ⇌ {OAc}^{-} (a q) + H_{3} O^{+}$ $"HOAc"(aq) + "H"_2"O"(l) rightleftharpoons "OAc"^(-)(aq) + "H"_3"O"^(+)$

$I 0.200 - 00$ $"I"" "" "0.200" "" "-" "" "" "0" "" "" "" "" "0$
$C - x - + x + x$ $"C"" "" "-x" "" "-" "" "" "+x" "" "" "" "+x$
$E 0.200 - x - x x$ $"E"" "0.200-x" "-" "" "" "x" "" "" "" "" "x$

The $K_{a}$ $K_a$ expression would then be:

$K_{a} = \frac{[{OAc}^{-}] [H_{3} O^{+}]}{[HOAc]}$ $K_a = (["OAc"^(-)]["H"_3"O"^(+)])/(["HOAc"])$

$= \frac{x^{2}}{0.200 - x}$ $= x^2/(0.200 - x)$

Since $K_{a}$ $K_a$ $\approx$ $~~$ $10^{- 5}$ $10^(-5)$ , we can use the small $x$ $x$ approximation to negligible error to get:

$K_{a} \approx \frac{x^{2}}{0.200} \Rightarrow x = \sqrt{0.200 K_{a}} = 0.001897 M$ $K_a ~~ x^2/(0.200) => x = sqrt(0.200K_a) = "0.001897 M"$

whereas the true $x$ $x$ would be from the quadratic formula on:

$x^{2} + K_{a} x - 0.200 K_{a} = 0$ $x^2 + K_ax - 0.200K_a = 0$

which would end up being $x = 0.001888 M$ $x = "0.001888 M"$ .

Nonetheless, we will proceed with the small $x$ $x$ approximation. From the Henderson-Hasselbalch equation, we can find the ratio of base to acid we need to generate for the buffer:

$"pH" = stackrel(-log(K_a))overbrace"pKa" + log\frac(["OAc"^(-)]_(buffer))(["HOAc"]_(buffer))$

$4.50 = -log(1.8 xx 10^(-5)) + log\frac(["OAc"^(-)]_(buffer))(["HOAc"]_(buffer))$

Thus, the ratio of concentrations of acetate to acetic acid is:

$(["OAc"^(-)]_(buffer))/(["HOAc"]_(buffer)) = 10^(4.50 - 4.7447) ~~ 0.5692_(099788)$

Since the $"mol"$ s of weak acid neutralized by $"NaOH"$ generates the same number of $"mol"$ s of its conjugate base, we can rewrite this ratio more conveniently as:

$(["OAc"^(-)]_(buffer))/(["HOAc"]_(buffer)) = (n_(OAc^(-))"/"V_(t ot) + n_(OH^(-))"/"V_(t ot))/(n_(HOAc)"/"V_(t ot) - n_(OH^(-))"/"V_(t ot))$

This is because the $"NaOH"$ , $"HOAc"$ , and $"OAc"^(-)$ share the same container. Thus, the total volume is shared and it can be cancelled out.

$=> (["OAc"^(-)]_(buffer))/(["HOAc"]_(buffer)) = (n_(OAc^(-)) + n_(OH^(-)))/(n_(HOAc) - n_(OH^(-))) = 0.5692$

So, the $"mol"$ s of acetate and acetic acid in the starting solution are:

$n_(OAc^(-)) = "0.001897 M" xx "0.5000 L" = 9.487 xx 10^(-4) "mols"$

$n_(HOAc) = (0.200 - 0.001897) "M" xx "0.5000 L" = "0.0991 mols"$

We do indeed know both before adding $"NaOH"$ . Thus, solve for $n_(OH^(-))$ :

$0.5692(n_(HOAc) - n_(OH^(-))) = n_(OAc^(-)) + n_(OH^(-))$

$0.5692n_(HOAc) - 0.5692n_(OH^(-)) = n_(OAc^(-)) + n_(OH^(-))$

$0.5692n_(HOAc) - n_(OAc^(-)) = 1.5692n_(OH^(-))$

$=> n_(OH^(-)) = (0.5692n_(HOAc) - n_(OAc^(-)))/(1.5692)$

$=$ $0.0353_(250487)$ $"mols NaOH"$

Without the small $x$ approximation we would get $n_(OH^(-)) = 0.0353_(279096)$ $"mols"$ still, which are pretty much the same. In the end, we get:

$color(blue)(V_(NaOH)) = n_("OH"^(-))/(["NaOH"]) = "0.01178 L"$

$=$ $color(blue)("11.78 mL NaOH")$

This would be the most exact answer using the most reasonable approximations.

METHOD 2

Ignoring the ICE table, we assume that:

$["HOAc"]_(eq) ~~ ["HOAc"]_i$

$n_(HOAc) ~~ "0.200 M" xx "0.500 L" = "0.100 mols"$

By ignoring the ICE table, we also assume that $n_(OAc^(-))$ $"<<"$ $n_("OH"^(-))$ . As a result, we can say the following:

$=> (["OAc"^(-)]_(buffer))/(["HOAc"]_(buffer)) ~~ (n_(OH^(-)))/(n_(HOAc) - n_(OH^(-))) = 0.5692$

Therefore, this method gives:

$0.5692n_(HOAc) - 0.5692n_(OH^(-)) = n_(OH^(-))$

$1.5692n_(OH^(-)) = 0.5692n_(HOAc)$

$n_(OH^(-)) ~~ (0.5692)/(1.5692)n_(HOAc)$

$~~ 0.0362_(7326)$ $"mols"$

So, the volume under this approximation that the weak acid dissociates negligibly is:

$color(blue)(V_(NaOH)) = "0.03627326 mols"/("3.0 M") ~~ "0.01209 L"$

$=$ $color(blue)("12.09 mL NaOH")$

This answer, from ignoring the ICE table, is off by a bit, but if you work it out, it's a $2.63%$ error, which isn't that bad.

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Suppose you have $500.0 mL$ $"500.0 mL"$ of $0.200 M$ $"0.200 M"$ acetic acid ( $K_{a} = 1.8 \times 10^{- 5}$ $K_a = 1.8 xx 10^(-5)$ ). To make a $pH$ $"pH"$ $4.50$ $4.50$ buffer, how many $mL$ $"mL"$ of $3.0 M$ $"3.0 M"$ $NaOH$ $"NaOH"$ must be added?

This is a fun challenge question this week in the chemistry class I'm TAing. I thought I might as well put it up here and answer it.

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This is a fun challenge question this week in the chemistry class I'm TAing. I thought I might as well put it up here and answer it.

1 Answer

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