Question #b17d5

1 Answer
Mar 10, 2016

Answer:

#K_a = 4.3 * 10^(-5)#

Explanation:

The first thing to notice here is that the pH of the resulting solution is lower than #7#. This should automatically tell you that the neutralization of the acid was not complete.

A complete neutralization of a weak acid with a strong base will produce a basic solution, since the conjugate base of the acid will react with water to reform some of the weak acid and produce hydroxide anions, #"OH"^(-)#, in solution.

An incomplete neutralization will result in the formation of a buffer solution that contains a weak acid, #"HA"#, and its conjugate base, #"A"^(-)#.

This means that the pH of the solution can be found using the Henderson - Hasselbalch equation

#color(blue)(|bar(ul(color(white)(a/a)"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))color(white)(a/a)|))#

Here

#color(purple)(|bar(ul(color(white)(a/a)color(black)(pK_a = -log(K_a))color(white)(a/a)|)))#

This is what you'll be solving for.

The balanced net ionic equation for this reaction looks like this

#"HA"_text((aq]) + "OH"_text((aq])^(-) -> "A"_text((Aq])^(-) + "H"_2"O"_text((l])#

Notice the #1:1# mole ratios that exist between the two reactants, on one hand, and between both reactants and the conjugate base #"A"^(-)#, on the other.

This tells you that every mole of weak acid that takes part in the reaction consumes one mole of sodium hydroxide and produces one mole of conjugate base #"A"^(-)#.

Use the molarities and volumes of the two solutions to figure out how many moles of each you're mixing

#color(blue)(|bar(ul(color(white)(a/a)c = n/V implies n = c * Vcolor(white)(a/a)|)))#

#n_(HA) = "0.20 mol" color(red)(cancel(color(black)("L"^(-1)))) * 35.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.00700 moles HA"#

#n_(OH^(-)) = "0.20 mol" color(red)(cancel(color(black)("L"^(-1)))) * 25.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.00500 moles OH"^(-)#

Since you have fewer moles of hydroxide anions (delivered to the solution by the sodium hydroxide), these will be consumed completely by the reaction.

According to the aforementioned mole ratios, #0.00500# moles of #"OH"^(-)# will consume #0.00500# moles of #"HA"# and produce #0.00500# moles of #"A"^(-)#.

This means that after the reaction is finished, the solution will contain

#n_(OH^(-)) = "0 moles" -> color(red)("completely"color(white)(a)color(black)("consumed"))#

#n_(HA) = 0.00700 - 0.00500 = "0.00200 moles HA"#

#n_(A^(-)) = 0 + 0.00500 = "0.00500 moles A"^(-)#

The total volume of the resulting solution will be

#V_"total" = "35.0 mL" + "25.0 mL" = "60.0 mL"#

The concentrations of the weak acid and of the conjugate base will be

#["HA"] = "0.00200 moles"/(60.0 * 10^(-3)"L") = "0.03333 M"#

#["A"^(-)] = "0.00500 moles"/(60.0 * 10^(-3)"L") = "0.08333 M"#

Before plugging this into the H - H equation, try to predict what the #pK_a# of the acid will be.

Notice that the pH is said o be equal to #4.77#, and that the buffer contains more conjugate base than weak acid. This should tell you that the #pK_a# of the acid is lower than #4.77#.

Rearrange the H - H equation to solve for #pK_a#

#pK_a = "pH" - log( (["A"^(-)])/(["HA"]))#

Plug in your values to get

#pK_a = 4.77 - log( (0.08333 color(red)(cancel(color(black)("M"))))/(0.03333color(red)(cancel(color(black)("M"))))) = 4.37#

As predicted, the #pK_a# of the acid is lower than #4.77#.

The acid dissociation constant, #K_a#, for this acid will be

#K_a = 10^(-pK_a) = 10^(-4.37) = color(green)(|bar(ul(color(white)(a/a)4.3 * 10^(-5)color(white)(a/a)|)))#

The answer is rounded to two sig figs.