# Question f0846

Nov 19, 2017

$\text{pH} = 8.313$

#### 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 $\to$ you can't use the Henderson - Hasselbalch equation to calculate the $\text{pH}$ of the solution.

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

${\text{HCOOH"_ ((aq)) + "OH"_ ((aq))^(-) -> "HCOO"_ ((aq))^(-) + "H"_ 2"O}}_{\left(l\right)}$

so when you're mixing equal volumes of $\text{0.150 M}$ sodium hydroxide solution and $\text{0.150 M}$ 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

${\text{0.0075 moles " - " 0.0075 moles" = "0 moles OH}}^{-}$

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

$\text{0.0075 moles " - " 0.0075 moles" = "0 moles HCOOH }$

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

and ${\text{0.0075 moles HCOO}}^{-}$ 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

${V}_{\text{total" = "50.00 mL + 50.00 mL}}$

${V}_{\text{total" = "100.00 mL}}$

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

$\left[\text{HCOO"^(-)] = "0.0075 moles"/(100.00 * 10^(-3)color(white)(.)"L}\right)$

["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.

${\text{HCOO"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCOOH"_ ((aq)) + "OH}}_{\left(a q\right)}^{-}$

You know that an aqueous solution at room temperature has

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{p"K_a + "p} {K}_{b} = 14}}}$

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

$\text{p} {K}_{b} = 14 - 3.75$

$\text{p} {K}_{b} = 10.25$

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 $\text{ " ["OH"^(-)] = xcolor(white)(.)"M}$

and

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

This basically tells you that in order for the solution to contain $x$ $\text{M}$ of formic acid and of hydroxide anions, the concentration of the formate anions must decrease by $x$ $\text{M}$.

By definition, the base dissociation constant is equal to

${K}_{b} = {10}^{- \text{p} {K}_{b}}$

and with

${K}_{b} = \left(\left[{\text{HCOOH"] * ["OH"^(-)])/(["HCOO}}^{-}\right]\right)$

This means that you have

${10}^{- 10.25} = \frac{x \cdot x}{0.075 \cdot x} = {x}^{2} / \left(0.075 - x\right)$

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

$0.075 - x \approx 0.075$

You will thus have

${10}^{- 10.25} = {x}^{2} / 0.075$

which gets you

$x = \sqrt{0.075 \cdot {10}^{- 10.25}} = 2.054 \cdot {10}^{- 6}$

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 $\text{pH}$ of the solution will be

$\text{pH} = 14 + \log \left(2.054 \cdot {10}^{- 6}\right)$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{pH} = 8.31}}}$

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

$\text{pH} = 8.313$

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 $\text{pH} > 7$.

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 $\text{pH}$ of the solution.

In other words, for a weak acid - strong base titration, the equivalence point is at $\text{pH} > 7$.