Question a3768

Dec 14, 2015

$\text{2.9 g}$

Explanation:

You're dealing with a buffer solution that contains formic acid, a weak acid, and sodium formate, its conjugate base.

As you know, the Henderson - Hasselbalch equation allows you to express the pH of the buffer in terms of the concentrations of the weak acid and the conjugate base and the $p {K}_{a}$ of the acid

color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))

You know that your buffer has a pH of $3.50$, and that the weak acid's $p {K}_{a}$ is equal to $3.74$.

This tells you that the buffer contains less conjugate base than weak acid, since the pH of the solution is slightly lower than the $p {K}_{a}$ of the acid.

Your goal now is to use the H - H equation to find the ratio that exists between the concentration of the conjugate base and that of the weak acid at that pH

$3.50 = 3.74 + \log \left(\left(\left[\text{HCOO"^(-)])/(["HCOOH}\right]\right)\right)$

$- 0.24 = \log \left(\left(\left[\text{HCOO"^(-)])/(["HCOOH}\right]\right)\right)$

This is equivalent to

${10}^{- 0.24} = {10}^{\log} \left(\left(\left[\text{HCOO"^(-)])/(["HCOOH}\right]\right)\right)$

$\left(\left[\text{HCOO"^(-)])/(["HCOOH}\right]\right) = {10}^{- 0.24} = 0.57544$

This means that the concentration of the conjugate base must be equal to

$\left[\text{HCOO"^(-)] = ["HCOOH}\right] \cdot 0.57544$

You know that the formic acid has a molarity of $\text{0.15 M}$, which means that you have

["HCOO"^(-)] = 0.57544 * "0.15 M" = "0.08632 M"

Use the volume of the buffer solution to determine how many moles of formate anions, ${\text{HCOO}}^{-}$, must be present in solution to produce that molarity.

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

$n = {\text{0.08632 M" * "0.5 L" = "0.04316 moles HCOO}}^{-}$

As you know, sodium formate dissociates in aqueous solution in a $1 : 1$ mole ratio to produce formate anions and sodium cations

${\text{HCOONa"_text((aq]) -> "HCOO"_text((aq])^(-) + "Na}}_{\textrm{\left(a q\right]}}^{+}$

This tells you that you need to have $0.04316$ moles of sodium formate to produce $0.04316$ moles of formate anions in solution.

Finally, use sodium formate's molar mass to determine how many grams of sodium formate would contain this many moles

0.04316 color(red)(cancel(color(black)("moles HCOONa"))) * "68.007 g"/(1color(red)(cancel(color(black)("mole HCOONa")))) = "2.9352 g"#

I'll leave the answer rounded to two sig figs, despite the fact thatyou only have one sig fig for the volume of the buffer

${m}_{H C O O N a} = \textcolor{g r e e n}{\text{2.9 g}}$