A 30.00 mL sample of 0.100 M "HOI" is titrated using 0.100 M "NaOH", K_a=2.3xx10^-11 for "HOI"?

A) write a balanced net-ionic equation for the titration reaction. B) calculate the pH of the titration mixture at the equivalence point. C) would methyl red be a suitable indicator for the titration?

Jul 30, 2018

A) Well, this first one should be the easiest part. The balanced molecular equation would be:

$\text{HOI"(aq) + "NaOH"(aq) -> "NaOI"(aq) + "H"_2"O} \left(l\right)$

Removing the spectator ion ${\text{Na}}^{+}$, we get:

$\textcolor{b l u e}{\text{HOI"(aq) + "OH"^(-)(aq) -> "OI"^(-)(aq) + "H"_2"O} \left(l\right)}$

Clearly, $\text{HOI}$ is a weak acid (what ${K}_{a}$ marks that boundary?), so we must stop here and NOT dissociate $\text{HOI}$ completely.

B) The equivalence point is something that has been addressed here, so you should go back here and review this (again).

We therefore know that $\text{30.00 mL}$ of $\text{0.100 M}$ $\text{NaOH}$ would be required here, and that the concentration you start with is ${\text{0.0500 M OI}}^{-}$ contained in $\text{60.00 mL}$ for the association in water, produced from the reaction of $\text{0.00300 mols HOI}$ with ${\text{0.00300 mols OH}}^{-}$:

((0.100 cancel"mol HOI")/cancel"L" xx 0.0300 cancel"L" xx ("1 mol OI"^(-))/cancel("1 mol HOI"))/("0.0300 L" + "0.0300 L") = "0.0500 M OI"^(-)

Being an anion, ${\text{OI}}^{-}$ is a weak conjugate base (of $\text{HOI}$), with

${K}_{b} = {K}_{w} / {K}_{a} = {10}^{- 14} / \left(2.3 \times {10}^{- 11}\right) = 4.35 \times {10}^{- 4}$

at ${25}^{\circ} \text{C}$. We must do this... ${K}_{a}$ is ONLY for acids. At this point we don't need an ICE table. You can make one if you want. We construct the mass action expression:

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

${K}_{b} = \left(\left[{\text{OH"^(-)]["HOI"])/(["OI}}^{-}\right]\right)$

$= {x}^{2} / \left(0.0500 - x\right)$

This ${K}_{b}$ is not small enough for the small $x$ approximation.

$4.35 \times {10}^{- 4} \approx {x}^{2} / 0.0500$

${x}_{1} \equiv \left[{\text{OH}}^{-}\right] = \sqrt{0.0500 {K}_{b}}$

$= 4.66 \times {10}^{- 3} \text{M}$

And this $x$ is not that small compared to ${\left[{\text{OI}}^{-}\right]}_{i}$ as can be seen below...

x_1/(["OI"^(-)]_i) xx 100% = 9.33%

${x}_{2} = \sqrt{\left(0.0500 - {x}_{1}\right) {K}_{b}} = \text{0.00444 M}$

${x}_{3} = \sqrt{\left(0.0500 - {x}_{2}\right) {K}_{b}} = \text{0.00445 M}$

${x}_{4} = \sqrt{\left(0.0500 - {x}_{3}\right) {K}_{b}} = \text{0.00445 M}$

and our answer has converged, showing that the percent dissociation is actually 8.90%.

Therefore, the $\text{pH}$ at ${25}^{\circ} \text{C}$ is:

color(blue)("pH") = 14 - "pOH"

$= 14 + \log \left[{\text{OH}}^{-}\right]$

$= \textcolor{b l u e}{11.65}$

This must mean that $\text{HOI}$ is a weaker acid than $\text{HCN}$. The $\text{pH}$ is higher at the equivalence point than $10.95$.

(Indeed it is, because ${K}_{a} \left(\text{HCN}\right) = 6.2 \times {10}^{- 10}$, so $\text{HCN}$ is the stronger weak acid.)

C) Well, what is the $\text{pH}$ range of methyl red? Apparently, it has a ${\text{pK}}_{a}$ of $5.1$, so it changes color near $\text{pH}$ $5.1$. Will methyl red work then? We wanted to observe the $\text{pH}$ change at the equivalence point, which is much larger than $5.1$... will we see a color change there?