# 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 #"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

The **equivalence point** is reached when mols of

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

#=# #"25 mL"#

**before** the equivalence point, we neutralize almost all *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"^(-)(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 **not** interfere):

#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