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.

1 Answer
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.

"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.


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.