Question

Question #a3768

Answer 1

`"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 `pK_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 `pK_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 `pK_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( (["HCOO"^(-)])/(["HCOOH"]))`

`-0.24 = log( (["HCOO"^(-)])/(["HCOOH"]))`

This is equivalent to

`10^(-0.24) = 10^log( (["HCOO"^(-)])/(["HCOOH"]))`

`(["HCOO"^(-)])/(["HCOOH"]) = 10^(-0.24) = 0.57544`

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

`["HCOO"^(-)] = ["HCOOH"] * 0.57544`

You know that the formic acid has a molarity of `"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, `"HCOO"^(-)`, must be present in solution to produce that molarity.

`color(blue)(c = n/V implies n = c * V)`

`n = "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

`"HCOONa"_text((aq]) -> "HCOO"_text((aq])^(-) + "Na"_text((aq])^(+)`

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_(HCOONa) = color(green)("2.9 g")`