# Feasibility of titration of "2.5 mmols HY" ("pK"_a = 7.00) using "0.1 mmol/mL" "NaOH"?

## Using the usual approximations, find the $\text{pH}$ if you are $\left(a\right)$ $\text{0.05 mL}$ before the equivalence point, $\left(b\right)$ at the equivalence point, and $\left(c\right)$ $\text{0.05 mL}$ past the equivalence point. Comment on the feasibility of the titration.

Jan 27, 2018

Well, we examine near the equivalence point, at the equivalence point, and after it. Le Chatelier shifts would then restrict us from getting near $9.70$.

$\text{pH} = 9.490 , 9.699 , 9.908$

I don't know what "usual approximations" you make, but had we ignored concentrations on the order of ${10}^{- 5} \text{M}$, we would then get $9.698 , 9.699$, and $9.698$. They can't be ignored. They are too close to the ${K}_{a}$ and ${K}_{b}$, so they interfere.

We have thus shown that the $\text{pH}$ is quite volatile this close to the equivalence point that if you stopped $\text{0.05 mL}$ before, you fall short by $\text{0.2 pH}$ units, and if you stopped $\text{0.05 mL}$ after, you get shot forward $\text{0.2 pH}$ units. (You DON'T want that!)

This makes sense, because the distance of each $\text{pH}$ from the equivalence point $\text{pH}$ is equal. The real values here have the odd function symmetry as required by the titration curve.

DISCLAIMER: UNCENSORED MATH!

By titrating the acid with $\text{NaOH}$, we generate a buffer, followed by just ${\text{Y}}^{-}$ as we keep adding $\text{NaOH}$.

The equivalence point is reached when mols of $\text{HY}$ equal mols of ${\text{OH}}^{-}$ added. Therefore, we want

${V}_{N a O H} = \text{2.5 mmol NaOH" cdot "1 mL"/"0.1 mmol}$

$=$ $\text{25 mL}$

a) If we are $\text{0.05 mL}$ before the equivalence point, we neutralize almost all $\text{HY}$, so that we are still in the buffer region.

The concentrations are:

["HY"] = ("2.5 mmols" - 24.95/25 cdot "2.5 mmols NaOH")/("75 mL" + "24.95 mL") = 5.00 xx 10^(-5) "M"

["Y"^(-)] = (24.95/25 cdot "2.5 mmols NaOH")/("75 mL" + "24.95 mL") = "0.02496 M"

And so, with ${\text{Y}}^{-}$ dominating, we construct the ICE table for base association:

${\text{Y"^(-)(aq) " "+" " "H"_2"O"(l) rightleftharpoons "HY"(aq) " "+" " "OH}}^{-} \left(a q\right)$

$\text{I"" "0.02496" "" "" "-" "" "" "5.00 xx 10^(-5)" "" } 0$
$\text{C"" "-x" "" "" "" "-" "" "" "" "+x" "" "" "" } + x$
$\text{E"" "0.02496-x" "-" "" "" "5.00 xx 10^(-5) + x" } x$

Since ${\text{pK}}_{a} = 7.00$, ${K}_{a} = {10}^{- 7}$, so ${K}_{b} = {10}^{- 7}$ at ${25}^{\circ} \text{C}$. Therefore:

${K}_{b} = {10}^{- 7} = \frac{\left(5.00 \times {10}^{- 5} + x\right) x}{0.02496 - x}$

Here we see that $x$ $\text{<<}$ $0.02496$, but not $5.00 \times {10}^{- 5}$, so

${10}^{- 7} \approx \frac{\left(5.00 \times {10}^{- 5} + x\right) x}{0.02496}$

${x}^{2} + 5.00 \times {10}^{- 5} x - {10}^{- 7} \cdot 0.02496 = 0$

which gives approximately $3.087 \times {10}^{- 5} \text{M}$ ${\text{OH}}^{-}$ (compared to the exact result of $3.084 \times {10}^{- 5} \text{M}$). As a result,

$\textcolor{b l u e}{\text{pH}} = - \log \left({10}^{- 14} / \left(3.087 \times {10}^{- 5}\right)\right) = \textcolor{b l u e}{9.490}$

and not close to $9.70$.

b) At the equivalence point we would then have zero $\text{HY}$ left and all ${\text{Y}}^{-}$. No $\text{NaOH}$ is present, and we let ${\text{Y}}^{-}$ freely associate in water.

${\text{Y"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HY"(aq) + "OH}}^{-} \left(a q\right)$

The concentration of ${\text{Y}}^{-}$ currently is

("2.5 mmols Y"^(-))/("75 mL soln" + "25 mL NaOH") = "0.025 M"

The ICE table becomes:

${\text{Y"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HY"(aq) + "OH}}^{-} \left(a q\right)$

$\text{I"" "0.025" "" "" "-" "" "0" "" "" "" } 0$
$\text{C"" "-x" "" "" "-" "" "+x" "" "" } + x$
$\text{E"" "0.025-x" "-" "" "x" "" "" "" } x$

Therefore:

${K}_{b} = {10}^{- 7} = {x}^{2} / \left(0.025 - x\right)$

Here the small $x$ approximation is appropriate; $K$ is quite small, so

${10}^{- 7} \approx {x}^{2} / 0.025$

and

$x \approx \sqrt{0.025 \cdot {10}^{- 7}} = 5 \times {10}^{- 5} {\text{M OH}}^{-}$

(and the true value is $4.995 \times {10}^{- 5} \text{M}$.)

So assuming that water autodissociation does not interfere since this is a factor of 100 larger,

$\textcolor{b l u e}{{\text{pH") = -log["H}}^{+}}$

$\approx - \log \left(\frac{{10}^{- 14}}{5 \times {10}^{- 5}}\right) = \textcolor{b l u e}{9.699}$

(If we had accounted for it, then we would get fortuitous cancellation that ["OH"^(-)] = 5.01 xx 10^(-5) "M" with no approximations, so that $\text{pH} = 9.699$ again.)

c) If we are $\text{0.05 mL}$ of $\text{NaOH}$ past the equivalence point, then we have some $\text{NaOH}$ interfering with the association of ${\text{Y}}^{-}$ in water via Le Chatelier's principle.

The new concentration of ${\text{Y}}^{-}$ is:

$\text{2.5 mmols"/("75 mL" + "25.05 mL") = "0.02499 M}$

The concentration of $\text{NaOH}$ present not from water is:

("0.05 mL" xx "0.1 mmols"/"mL")/("75 mL" + "25.05 mL") = 4.998 xx 10^(-5) "M"

${\text{Y"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HY"(aq) + "OH}}^{-} \left(a q\right)$

$\text{I"" "0.02499" "" "" "-" "0" "" "" "" } 4.998 \times {10}^{- 5}$
$\text{C"" "-x" "" "" "" "-" "+x" "" "" } + x$
$\text{E"" "0.02499-x" "-" "x" "" "" "" } 4.998 \times {10}^{- 5} + x$

And so

${K}_{b} = {10}^{- 7} = \frac{\left(4.998 \times {10}^{- 5} + x\right) x}{0.02499 - x}$

Here we recognize that $x$ $\text{<<}$ $0.02499$ but not $4.998 \times {10}^{- 5}$, so we get the approximate quadratic equation (assuming water autodissociation does not interfere):

${x}^{2} + 4.998 \times {10}^{- 5} x - {10}^{- 7} \cdot 0.02499 = 0$

The approximate solution is $x = 3.099 \times {10}^{- 5} \text{M}$ (compared to $3.087 \times {10}^{- 5} \text{M}$), so

["OH"^(-)] = (4.998 + 3.099) xx 10^(-5) "M"

$= 8.097 \times {10}^{- 5} \text{M}$

Therefore, the $\text{pH}$ with these approximations is:

$\textcolor{b l u e}{{\text{pH") = 14 - "pOH" = 14 + log["OH}}^{-}}$

$= \textcolor{b l u e}{9.908}$

If we allowed water to interfere in our assumptions, and then also took the exact solution to the previous quadratic, then $\left[{\text{OH}}^{-}\right]$ turns out to be $8.095 \times {10}^{- 5} \text{M}$ due to fortuitous cancellation, and the actual $\text{pH}$ would be more like $9.909$.