HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)

A student was given the task of titrating a 20.mL sample of 0.10MHCl(aq) with 0.10MNaOH(aq). The HCl(aq) was placed in an Erlenmeyer flask. An equation for the reaction that occurs during the titration is given above. (a) According to the equation for the reaction, if the amount of the reactants is halved, how does this affect the amount of H2O(l) produced in the reaction?

According to the reaction equation shown,  if the amount of the reactants is halved, the amount of H2O(l) produced in the reaction will be halved.

The molecular reaction equation was given as follows; HCl(aq) + NaOH(aq)→ NaCl(aq) + H2O(l). Ionically, we can write; H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) > Na^+(aq) + + Cl^-(aq) + H2O(l). The net ionic equation now becomes; H^+(aq) + OH^-(aq) > H2O(l).

This implies that the concentration of H2O(l) depends on the concentrations of H^+(aq) and OH^-(aq) from HCl(aq) and NaOH(aq) respectively and the mole ratio of reactants and products is 1:1:1:1.

Hence, of the amount of the reactants is halved,  the amount of H2O(l) produced in the reaction will be halved.

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