Question #a3768
1 Answer
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
#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))#
You know that your buffer has a pH of
This tells you that the buffer contains less conjugate base than weak acid, since the pH of the solution is slightly lower than the
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
#["HCOO"^(-)] = 0.57544 * "0.15 M" = "0.08632 M"#
Use the volume of the buffer solution to determine how many moles of formate anions,
#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
#"HCOONa"_text((aq]) -> "HCOO"_text((aq])^(-) + "Na"_text((aq])^(+)#
This tells you that you need to have
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")#