A 30.00 mL sample of 0.100 M $HOI$ $"HOI"$ is titrated using 0.100 M $NaOH$ $"NaOH"$ , $K_{a} = 2.3 \times 10^{- 11}$ $K_a=2.3xx10^-11$ for $HOI$ $"HOI"$ ?

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A 30.00 mL sample of 0.100 M $HOI$ $"HOI"$ is titrated using 0.100 M $NaOH$ $"NaOH"$ , $K_{a} = 2.3 \times 10^{- 11}$ $K_a=2.3xx10^-11$ for $HOI$ $"HOI"$ ?

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?

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Answer 1 · 2018-07-30T16:53:51

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

$HOI (a q) + NaOH (a q) \to NaOI (a q) + H_{2} O (l)$ $"HOI"(aq) + "NaOH"(aq) -> "NaOI"(aq) + "H"_2"O"(l)$

Removing the spectator ion ${Na}^{+}$ $"Na"^(+)$ , we get:

$HOI (a q) + {OH}^{-} (a q) \to {OI}^{-} (a q) + H_{2} O (l)$ $color(blue)("HOI"(aq) + "OH"^(-)(aq) -> "OI"^(-)(aq) + "H"_2"O"(l))$

Clearly, $HOI$ $"HOI"$ is a weak acid (what $K_{a}$ $K_a$ marks that boundary?), so we must stop here and NOT dissociate $HOI$ $"HOI"$ completely.

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

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

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

Being an anion, $"OI"^(-)$ is a weak conjugate base (of $"HOI"$ ), with

$K_b = K_w/K_a = 10^(-14)/(2.3 xx 10^(-11)) = 4.35 xx 10^(-4)$

at $25^@ "C"$ . 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:

$"OI"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "OH"^(-)(aq) + "HOI"(aq)$

$K_b = (["OH"^(-)]["HOI"])/(["OI"^(-)])$

$= x^2/(0.0500 - x)$

This $K_b$ is not small enough for the small $x$ approximation.

$4.35 xx 10^(-4) ~~ x^2/0.0500$

$x_1 -= ["OH"^(-)] = sqrt(0.0500K_b)$

$= 4.66 xx 10^(-3) "M"$

And this $x$ is not that small compared to $["OI"^(-)]_i$ as can be seen below...

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

So, we adjust accordingly.

$x_2 = sqrt((0.0500 - x_1)K_b) = "0.00444 M"$

$x_3 = sqrt((0.0500 - x_2)K_b) = "0.00445 M"$

$x_4 = sqrt((0.0500 - x_3)K_b) = "0.00445 M"$

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

Therefore, the $"pH"$ at $25^@ "C"$ is:

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

$= 14 + log["OH"^(-)]$

$= color(blue)(11.65)$

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

(Indeed it is, because $K_a("HCN") = 6.2 xx 10^(-10)$ , so $"HCN"$ is the stronger weak acid.)

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

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A 30.00 mL sample of 0.100 M $HOI$ $"HOI"$ is titrated using 0.100 M $NaOH$ $"NaOH"$ , $K_{a} = 2.3 \times 10^{- 11}$ $K_a=2.3xx10^-11$ for $HOI$ $"HOI"$ ?

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?

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A 30.00 mL sample of 0.100 M "HOI"HOI"HOI" is titrated using 0.100 M "NaOH"NaOH"NaOH", K_a=2.3xx10^-11Ka=2.3×10−11K_a=2.3xx10^-11 for "HOI"HOI"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?

1 Answer

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Impact of this question

A 30.00 mL sample of 0.100 M $HOI$ $"HOI"$ is titrated using 0.100 M $NaOH$ $"NaOH"$ , $K_{a} = 2.3 \times 10^{- 11}$ $K_a=2.3xx10^-11$ for $HOI$ $"HOI"$ ?

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?