Question

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?

Answer 1

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

`"HOI"(aq) + "NaOH"(aq) -> "NaOI"(aq) + "H"_2"O"(l)`

Removing the spectator ion `"Na"^(+)`, we get:

`color(blue)("HOI"(aq) + "OH"^(-)(aq) -> "OI"^(-)(aq) + "H"_2"O"(l))`

Clearly, `"HOI"` is a weak acid (what `K_a` marks that boundary?), so we must stop here and NOT dissociate `"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 `"30.00 mL"` of `"0.100 M"` `"NaOH"` would be required here, and that the concentration you start with is `"0.0500 M OI"^(-)` contained in `"60.00 mL"` for the association in water, produced from the reaction of `"0.00300 mols HOI"` with `"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?