Question

Question #f0846

Answer 1

`"pH" = 8.313`

Explanation:

!! LONG ANSWER !!

The trick here is to realize that you're no longer dealing with a buffer solution because all the moles of formic acid have been neutralized.

So if your solution doesn't contain a weak acid and its conjugate base in comparable amounts (or a weak base and its conjugate acid), then it's not a buffer `->` you can't use the Henderson - Hasselbalch equation to calculate the `"pH"` of the solution.

You know that formic acid and sodium hydroxide react in a `1:1` mole ratio

`"HCOOH"_ ((aq)) + "OH"_ ((aq))^(-) -> "HCOO"_ ((aq))^(-) + "H"_ 2"O"_ ((l))`

so when you're mixing equal volumes of `"0.150 M"` sodium hydroxide solution and `"0.150 M"` formic acid solution, you're mixing equal numbers of moles of each reactant.

More specifically, you're mixing

`50.00 color(red)(cancel(color(black)("mL solution"))) * "0.150 moles"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0075 moles"`

of hydroxide anions and of formic acid. Since the reaction produces formate anions in a `1:1` mole ratio with both reactants, you can say that after the reaction is complete, your solution will contain

`"0.0075 moles " - " 0.0075 moles" = "0 moles OH"^(-)`

All the moles of hydroxide anions present in the solution are consumed!

`"0.0075 moles " - " 0.0075 moles" = "0 moles HCOOH "`

All the moles of formic acid present in the formic acid solution are consumed!

and `"0.0075 moles HCOO"^(-)` because the reaction consumes `0.0075` moles of both reactants and produces `0.0075` moles of formate anions.

The total volume of the solution will be

`V_"total" = "50.00 mL + 50.00 mL"`

`V_"total" = "100.00 mL"`

The concentration of the formate anions in the resulting solution will be

`["HCOO"^(-)] = "0.0075 moles"/(100.00 * 10^(-3)color(white)(.)"L")`

`["HCOO"^(-)] = "0.075 M"`

So now your solution contains the formate anion, the conjugate base of the weak acid. The formate anions will act as a base in aqueous solution to produce formic acid and hydroxide anions.

`"HCOO"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCOOH"_ ((aq)) + "OH"_ ((aq))^(-)`

You know that an aqueous solution at room temperature has

`color(blue)(ul(color(black)("p"K_a + "p"K_b = 14)))`

so you can say that the base dissociation constant for the formate anions is equal to

`"p"K_b = 14 - 3.75`

`"p"K_b = 10.25`

If you take `x` to be the equilibrium concentration of formic acid and of hydroxide anions--note that the formate anions ionize in a `1:1` mole ratio to produce formic acid and hydroxide anions--you can say that your solution will contain

`["HCOOH"] = xcolor(white)(.)"M" " "` and `" " ["OH"^(-)] = xcolor(white)(.)"M"`

and

`["HCOO"^(-)] = (0.075 - x)color(white)(.)"M"`

This basically tells you that in order for the solution to contain `x` `"M"` of formic acid and of hydroxide anions, the concentration of the formate anions must decrease by `x` `"M"`.

By definition, the base dissociation constant is equal to

`K_b = 10^(-"p"K_b)`

and with

`K_b = (["HCOOH"] * ["OH"^(-)])/(["HCOO"^(-)])`

This means that you have

`10^(-10.25) = (x * x)/(0.075 * x) = x^2/(0.075 - x)`

Now, because the base dissociation constant is so small compared to the initial concentration of the formate anions, you can use the approximation

`0.075 - x ~~ 0.075`

You will thus have

`10^(-10.25) = x^2/0.075`

which gets you

`x = sqrt(0.075 * 10^(-10.25)) = 2.054 * 10^(-6)`

This means that, at equilibrium, the resulting solution will contain

`["OH"^(-)] = 2.054 * 10^(-6)color(white)(.)"M"`

As you know, an aqueous solution at room temperature has

`color(blue)(ul(color(black)("pH" = 14 + log(["OH"^(-)]))))`

You can thus say that the `"pH"` of the solution will be

`"pH" = 14 + log(2.054 * 10^(-6))`

`color(darkgreen)(ul(color(black)("pH" = 8.31)))`

I'll leave the answer rounded to two decimal places, but you could report it as

`"pH" = 8.313`

because you have three sig figs for the concentration of the two solutions.

Finally, does this result make sense?

Right from the start, the fact that you're using a strong base to neutralize a weak acid should tell you that the resulting solution must have `"pH" > 7`.

This is the case because the conjugate base of the weak acid will ionize to reform some of the weak acid and to produce hydroxide anions, which increase the `"pH"` of the solution.

In other words, for a weak acid - strong base titration, the equivalence point is at `"pH" > 7`.

https://www.khanacademy.org/test-prep/mcat/chemical-processes/titrations-and-solubility-equilibria/a/acid-base-titration-curves