A solution of 0.100 M NaOH is the titrant added from a burette. 25 mL of a 0.100

Question

A solution of 0.100 M NaOH is the titrant added from a burette. 25 mL of a 0.100 M HCOOH is the analyte.

a) Calculate the concentration of the formate ion HCOO- at the equivalence point. Is it acidic, neutral or basic? Calculate pH. (hint what is Kb for the formate ion?)

b)Based on your answer above what volume of NaOH is required to reach half the equivalence point? determine the concentration of HCOOH and HCOO- at the equivalence point.

c) What is the pH limit that the reaction solution approaches as more NaOH is added from the burette after the equivalence point has been reached?

Explanation / Answer

The reaction of NaOH and HCOOH is a 1 : a mole ratio basis

HCOOH + NaOH <==> HCOONa + H2O

initial moles of HCOOH = molarity x volume = 0.1 M x 0.025 L = 0.0025 moles

It would react with 0.0025 moles of NaOH to give 0.0025 moles of HCOONa

Volume of NaOH = moles/molarity = 0.0025/0.1 = 0.025 L = 25 ml

total volume = 25 + 25 = 50 ml = 0.050 L

molarity of HCOONa = moles/L = 0.0025/0.05 = 0.05 M

a) at equivalence point moles of NaOH = moles of HCOOH

that is, 0.0025 moles NaOH = 0.0025 moles HCOOH = 0.0025 moles of HCOONa

we have already calculated Molar concentration,

[HCOO-] = 0.05 M

the salt hydrolyzes as,

HCOO- + H2O <==> HCOOH + OH-

Kb = [HCOOH][OH-]/[HCOO-]

let x be the amount that has hydrolyzed then,

Ka for HCOOH = 1.77 x 10^-4

Kb = Kw/Ka = 1 x 10^-14/1.77 x 10^-4 = 5.65 x 10^-11

Kb = 5.65 x 10^-11 = x^2/0.05

x = [OH-] = 1.68 x 10^-6 M

pOH = -log[OH-] = -log(1.68 x 10^-6) = 5.77

pH = 14 - pH = 14 - 5.77 = 8.23

So the solution is basic at equivalence point

b) To reach half the equivalence point we would require

0.0025 moles HCOO- at equivalence point

So, 0.00125 moles at half equivalence point

volume of NaOH = 0.00125/0.1 = 0.0125 L = 12.5 mL

c) As more NaOH is added

pH = 14 - pOH

NaOH added dissociates to Na+ and OH-

So, pOH = -log[moles of excess NaOH added/total volume]

From this we calculate pH = 14 - pOH will be the limit.

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