The molecular formula of malonic acid is C3H4O4. If a solution containing 0.762 g of the acid requires 12.44 ml of 1.174 M NaOH for neutralization, how many ionizable H atoms are present in the molecule? Then write the molecular, total ionic, and net ionic equations for the reaction.

 

Answer:

From the given volume of a NaOH solution needed to neutralize the acid solution we determine the moles of NaOH needed and compare that with the moles of acid.

The number of moles present in 12.44 mL of 1.174 M NaOH = 12.44/1000 x 1.174 mol NaOH = 0.01460 mol NaOH

The number of moles present in 0.762 g malonic acid, C3H4O4 = 0.762g/104.06g/mol

= 0.00732 mol

It can be seen that twice as many moles of NaOH were needed to neutralize the malonic acid. This

means that malonic acid has two ionizable protons

Molecular

C₃H₄O₄(aq) + 2NaOH(aq) --------> Na₂C₃H₂O₄ (aq) + 2H₂O(l)

total ionic

2H⁺ (aq) + C3H2O4 ^2-(aq) + 2Na⁺(aq) + 2OH^-(aq) -------->C3H2O4^2-(aq) + 2Na+(aq) + 2H2O(l)

net ionic

H⁺(aq) + OH^-(aq) --------------> H₂O (l)