Question

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

Using the usual approximations, find the `"pH"` if you are `(a)` `"0.05 mL"` before the equivalence point, `(b)` at the equivalence point, and `(c)` `"0.05 mL"` past the equivalence point. Comment on the feasibility of the titration.

Answer 1

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`.

`"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) "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 `"pH"` is quite volatile this close to the equivalence point that if you stopped `"0.05 mL"` before, you fall short by `"0.2 pH"` units, and if you stopped `"0.05 mL"` after, you get shot forward `"0.2 pH"` units. (You DON'T want that!)

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

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DISCLAIMER: UNCENSORED MATH!

By titrating the acid with `"NaOH"`, we generate a buffer, followed by just `"Y"^(-)` as we keep adding `"NaOH"`.

The equivalence point is reached when mols of `"HY"` equal mols of `"OH"^(-)` added. Therefore, we want

`V_(NaOH) = "2.5 mmol NaOH" cdot "1 mL"/"0.1 mmol"`

`=` `"25 mL"`

`a)` If we are `"0.05 mL"` before the equivalence point, we neutralize almost all `"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 `"Y"^(-)` dominating, we construct the ICE table for base association:

`"Y"^(-)(aq) " "+" " "H"_2"O"(l) rightleftharpoons "HY"(aq) " "+" " "OH"^(-)(aq)`

`"I"" "0.02496" "" "" "-" "" "" "5.00 xx 10^(-5)" "" "0`
`"C"" "-x" "" "" "" "-" "" "" "" "+x" "" "" "" "+x`
`"E"" "0.02496-x" "-" "" "" "5.00 xx 10^(-5) + x" "x`

Since `"pK"_a = 7.00`, `K_a = 10^(-7)`, so `K_b = 10^(-7)` at `25^@ "C"`. Therefore:

`K_b = 10^(-7) = ((5.00 xx 10^(-5) + x)x)/(0.02496 - x)`

Here we see that `x` `"<<"` `0.02496`, but not `5.00 xx 10^(-5)`, so

`10^(-7) ~~ ((5.00 xx 10^(-5) + x)x)/0.02496`

So, we have the quadratic:

`x^2 + 5.00 xx 10^(-5)x - 10^(-7) cdot 0.02496 = 0`

which gives approximately `3.087 xx 10^(-5) "M"` `"OH"^(-)` (compared to the exact result of `3.084 xx 10^(-5) "M"`). As a result,

`color(blue)("pH") = -log(10^(-14)/(3.087 xx 10^(-5))) = color(blue)(9.490)`

and not close to `9.70`.

`b)` At the equivalence point we would then have zero `"HY"` left and all `"Y"^(-)`. No `"NaOH"` is present, and we let `"Y"^(-)` freely associate in water.

`"Y"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HY"(aq) + "OH"^(-)(aq)`

The concentration of `"Y"^(-)` currently is

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

The ICE table becomes:

`"Y"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HY"(aq) + "OH"^(-)(aq)`

`"I"" "0.025" "" "" "-" "" "0" "" "" "" "0`
`"C"" "-x" "" "" "-" "" "+x" "" "" "+x`
`"E"" "0.025-x" "-" "" "x" "" "" "" "x`

Therefore:

`K_b = 10^(-7) = x^2/(0.025 - x)`

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

`10^(-7) ~~ x^2/0.025`

and

`x ~~ sqrt(0.025 cdot 10^(-7)) = 5 xx 10^(-5) "M OH"^(-)`

(and the true value is `4.995 xx 10^(-5) "M"`.)

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

`color(blue)("pH") = -log["H"^(+)]`

`~~ -log((10^(-14))/(5 xx 10^(-5))) = color(blue)(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 `"pH" = 9.699` again.)

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

The new concentration of `"Y"^(-)` is:

`"2.5 mmols"/("75 mL" + "25.05 mL") = "0.02499 M"`

The concentration of `"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"`

`"Y"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HY"(aq) + "OH"^(-)(aq)`

`"I"" "0.02499" "" "" "-" "0" "" "" "" "4.998 xx 10^(-5)`
`"C"" "-x" "" "" "" "-" "+x" "" "" "+x`
`"E"" "0.02499-x" "-" "x" "" "" "" "4.998 xx 10^(-5) + x`

And so

`K_b = 10^(-7) = ((4.998 xx 10^(-5) + x)x)/(0.02499 - x)`

Here we recognize that `x` `"<<"` `0.02499` but not `4.998 xx 10^(-5)`, so we get the approximate quadratic equation (assuming water autodissociation does not interfere):

`x^2 + 4.998 xx 10^(-5)x - 10^(-7) cdot 0.02499 = 0`

The approximate solution is `x = 3.099 xx 10^(-5) "M"` (compared to `3.087 xx 10^(-5) "M"`), so

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

`= 8.097 xx 10^(-5) "M"`

Therefore, the `"pH"` with these approximations is:

`color(blue)("pH") = 14 - "pOH" = 14 + log["OH"^(-)]`

`= color(blue)(9.908)`

If we allowed water to interfere in our assumptions, and then also took the exact solution to the previous quadratic, then `["OH"^(-)]` turns out to be `8.095 xx 10^(-5) "M"` due to fortuitous cancellation, and the actual `"pH"` would be more like `9.909`.