Question #b17d5

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.