Question

160 g of a 0.80% sodium hydroxide solution.

The following question involves 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) + H₂C₄H₄O₄ (aq) --> Na₂C₄H₄O₄ (aq) + 2 H₂O (l)

A student weighs 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 A 25.00 mL sample of this solution is withdrawn and placed in a 125 mL Erlenmeyer flask, to which 5 drops of the acid-base indicator phenolphthalein is added.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?

0.0572 M

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

1.43 x 10^-3 mol

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:

14.30 mL NaOH

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.

0.214 M NaOH

You are going to standardize a 0.80% solution of sodium hydroxide solution for which the molarity is approximately 0.20 M. You will have dry succinic acid, a 250 mL volumetric flask, a 25.00 mL pipet, a 50.00 mL buret, a phenolphthalein indicator solution, and other standard laboratory equipment. For each titration you will withdraw a 25.00 mL sample of the succinic acid solution with a volumetric pipet. Therefore, you will deliver exactly one-tenth of your initial mass of succinic acid (to four significant figures) in each titration. In deciding the amount of succinic acid to weigh, assume that you will be using approximately 15 mL of the sodium hydroxide of approximate 0.20 M concentration for each titration, and you need enough succinic acid for ten titrations. Show how to calculate the mass of succinic acid you will need to weigh.

Answer 1

Volume of NaOH used = 15 mL.

Approx molarity of NaOH used = 0.20 M.

Approx. moles of NaOH used in each titration = (15 mL)*(1 L)/(1000 mL)*(0.20 M) = 0.0030 mole.

The balanced chemical equation for the reaction is given. As per the stoichiometric equation,

1 mole succinic acid = 2 moles NaOH.

Therefore,

moles succinic acid corresponding to 0.0030 mole = (0.0030 mole NaOH)*(1 mole succinic acid)/(2 moles of NaOH) = 0.0015 mole succinic acid.

The gram molar mass of succinic acid, H₂C₄H₄O₄ = (4*12.011 + 6*1.008 + 4*15.999) g/mol = 118.088 g/mol.

Mass of succinic acid required for each trial = (moles of succinic acid)*(molar mass of succinic acid) = (0.0015 mole)*(118.088 g/mol) = 0.177132 g.

Note that 10 titrations will be performed and each titration requires 0.177132 g succinic acid. Therefore, the mass of succinic acid required for 10 titrations = (0.177132 g)*10 = 1.77132 g ≈ 1.771 g (ans).