Question

A 10.00 mL sample of a solution containing formic acid (a weak acid) was placed in a 25 mL volumetric flask and diluted to the mark with water. A 10.00 mL sample of the diluted formic acid solution was then titrated with 0.1322 M sodium hydroxide. The titration required 15.80 mL of sodium hydroxide to reach the equivalence point. Calculate the molarity and the percentage (by mass) formic acid in the original solution. The density of the formic acid solution was found to be 1.02 g/mL.

Answer 1

Answer:

Molarity: 0.522M

Percentage by mass: 2.36 (w/w) %

Explanation:

Formic acid, HCOOH reacts with NaOH as follows:

HCOOH + NaOH → NaCOOH + H₂O

To solve this question we must find the moles of NaOH added = Moles formic acid. Taken into account the dilution that was made we can find the moles -And molarity of formic acid and its percentage by mass as follows:

Moles NaOH = Moles HCOOH:

0.01580L * (0.1322mol / L) =0.002089 moles HCOOH

Moles in the original solution:

0.002089 moles HCOOH * (25mL / 10mL) = 0.005222 moles HCOOH

Molarity of the solution:

0.005222 moles HCOOH / 0.01000L =

0.522M

Mass HCOOH in 1L -Molar mass: 46.03g/mol-

0.522moles * (46.03g / mol) = 24.04g HCOOH

Mass solution:

1L = 1000mL * (1.02g / mL) = 1020g solution

Mass percent:

24.04g HCOOH / 1020g solution * 100

2.36 (w/w) %