Question #a3768

1 Answer
Dec 14, 2015

Answer:

#"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")#