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.
Using the usual approximations, find the
1 Answer
Well, we examine near the equivalence point, at the equivalence point, and after it. Le Chatelier shifts would then restrict us from getting near
"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
We have thus shown that the
This makes sense, because the distance of each
DISCLAIMER: UNCENSORED MATH!
By titrating the acid with
The equivalence point is reached when mols of
V_(NaOH) = "2.5 mmol NaOH" cdot "1 mL"/"0.1 mmol"
= "25 mL"
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"^(-)(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
K_b = 10^(-7) = ((5.00 xx 10^(-5) + x)x)/(0.02496 - x)
Here we see that
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
color(blue)("pH") = -log(10^(-14)/(3.087 xx 10^(-5))) = color(blue)(9.490) and not close to
9.70 .
"Y"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HY"(aq) + "OH"^(-)(aq)
The concentration of
("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
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.)
The new concentration of
"2.5 mmols"/("75 mL" + "25.05 mL") = "0.02499 M"
The concentration of
("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^2 + 4.998 xx 10^(-5)x - 10^(-7) cdot 0.02499 = 0
The approximate solution is
["OH"^(-)] = (4.998 + 3.099) xx 10^(-5) "M"
= 8.097 xx 10^(-5) "M"
Therefore, the
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