Question

The following questions involve the reaction of sodium hydroxide and succinic acid. Succinic acid is diprotic: the acid (H₂C₄H₄O₄) reacts with two of its hydrogens to give C₄H₄O₄^2-. The balanced "molecular" equation for the reaction has coefficients of 2 for NaOH and 1 for succinic acid:

2 NaOH (aq) + H₂C₄O₄ (aq) --> Na₂C₄H₄O₄ (aq) + 2 H₂O (l)

A student weights 1.700 g of succinic acid and dissolves it in water in a 250.0 mL volumetric flask. A 25.00 mL sample of this solution is withdrawn and placed in a 125 mL Erlenmeyer flask, to which 5 drops of the acid-base indicator phenolphthalein is added. This solution is titrated with a sodium hydroxide solution of unknown molarity in a buret.

a) How many moles of succinic acid were weighed and dissolved in the volumetric flask?

b) How many moles of succinic acid were placed in the Erlenmeyer flask?

c) If the unknown sodium hydroxide solution was known to be approximately 0.20 M, at approximately what buret volume would you expect the end point of the titration? Show the calculation, including the use of the appropriate coefficients, as a 2 significant figure estimate.

d) If the measured volume at the end point was 13.32 mL of sodium hydroxide added, calculate the actual molarity of the sodium hydroxide solution.

Answer 1

a)

Mol of succinic acid = mass / molar mass

= 1.7 g / 118.09 gmol-1

= 0.0143 mol

Molarity = mol / volume of solution in litres

= 0.0143 mol / 0.25 L

= 0.0572 M

b)

mol = Molarity x volume

= 0.0572 M x 0.025 L

= 1.43 x 10^-3 mol

c)

As we can see from the chemical equation, it requires 2 mol of NaOH to neutralize 1 mol of succinic acid.

So,

2 x M1V1 = M2V2

Where, M1, M2 = Molarities of succinic acid and NaOH respectively

V1, V2 = Volumes of succinic acid and NaOH respectively

2 x 0.0572 M x 25 mL = 0.2 M x V2

14.3 mL = V2 = Volume of NaOH

d)

Similarly,

2 x 0.0572 x 25 mL = M2 x 13.32

0.214 M = M2 = Molarity of NaOH