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

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#.