Question #0760a

!! LONG ANSWER !!

Before doing anycalculations, make sure that you have a clear understanding of what goes on here.

You're titrating acetic acid, #"CH"""_3"COOH"#, a weak acid, with sodium hydroxide, #"NaOH"#, a strong base. Right from the start, this tells you that the pH of the solution at the equivalence point will not be equal to #7#.

That happens because the weak acid will react with the strong base to form sodium acetate, #"CH"""_3"COONa"#, which exists as sodium cations and acetate anions, #"CH"""_3"COO"^(-)#, in solution.

The acetate ion will then react with water to reform some acetic acid, producing hydroxide ions in the process.

This lets you know that the pH of the solution at equivalence point will be bigger than #7#.

So, determine the volume of sodium hydroxide solution that you must add to reach equivalence point using the mole ratio that exists between sodium hydroxide and acetic acid

#"CH"""_3"COOH"_text((aq]) + "NaOH"_text((aq]) -> "CH"""_3"COONa"_text((aq]) + "H"_2"O"_text((l])#

Every mole of acetic acid requires 1 mole of sodium hydroxide and produces 1 mole of acetate ions.

#C = n/V implies n = C * V#

#n_"acetic ac." = "0.100 M" * 25.00 * 10^(-3)"L" = "0.0025 moles"#

At the equivalence point, you have

#n_"NaOH" = n_"acetic ac." = "0.0025 moles"#

This means that you must add

#V = n/C = "0.0025 moles"/"0.125 M" = "20.00 mL NaOH"#

After the equivalence point is reached, your solution will have a total volume of

#V_"total" = V_"acetic ac." + V_(NaOH)#

#V_"total" = 25.00 + 20.00 = "45.00 mL"#

and contain 0.0025 moles of acetate ions, which were porduced by the neutralization reaction in accordance to the #1:1# mole ratio that exists between all the species involved.

The molarity of the acetate ions will be

#["CH"_3"COO"^(-)] = "0.0025 moles"/(45.00 * 10^(-3)"L") = "0.05556 M"#

Use an ICE table to find the concentration of the hydroxide ions - all the species are in aqueous solution (except water), so I won't add the states for simplicity

#"CH"_3"COO"^(-) + "H"_2"O" " "rightleftharpoons" " "CH"_3"COOH" + "OH"^(-)#

#color(purple)("I")" " "0.05556" " " " " " " " " " " " " " " " " " " 0" " " " " " " " " 0#
#color(purple)("C")" " ""(-x)" " " " " " " " " " " " " " " " " " "(+x)" " " " " " "(+x)#
#color(purple)("E")" " ""0.05556-x" " " " " " " " " " " " " " ""x" " " " " " " " " x#

Use the base dissociation constant, #K_b#, which you can find from

#pK_b = 14 - pK_a = 14 - 4.76 = 9.24#

#K_b = 10^(-pK_b) = 10^(-9.24) = 5.75 * 10^(-10)#

By definition, #K_b# is equal to

#K_b = (["CH"_3"COOH"] * ["OH"^(-)])/(["CH"_3"COO"^(-)]) = (x * x)/(0.05556 - x)#

Because #K_b# is so small, you can approximate #(0.05556 - x)# with #0.05556#. This will get you

#K_b = x^2/0.05556 implies x = sqrt(0.05556 * 5.75 * 10^(-10))#

#x = 5.65 * 10^(-6)#

This means that you have

#x = ["OH"^(-)] = 5.65 * 10^(-6)"M"#

The #pOH# of the solution will be

#pOH = -log(["OH"^(-)]) = -log(5.65 *10^(-6)) = 5.248#

The pH of the solution will thus be

#pH_"sol" = 14 - pOH = 14 - 5.248 = color(green)(8.752)#

As predicted, the pH of the solution is indeed bigger than #7#.