Question

The reaction of HCl with NaOH is represented by the equation HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) What volume of 0.6310 M HCl is required to titrate 15.80 mL of 0.3210 M NaOH?

Answer 1

This is a stoichiometry problem involving an acid-base titration. The balanced chemical equation for the reaction is:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

The stoichiometric coefficients indicate that one mole of HCl reacts with one mole of NaOH. Therefore, we can determine the number of moles of HCl required to react with 0.3210 M NaOH:

0.3210 mol/L NaOH × 0.01580 L NaOH = 0.00507 mol NaOH

Since the mole ratio of NaOH to HCl is 1:1, we need 0.00507 moles of HCl to react with the NaOH. To calculate the volume of 0.6310 M HCl needed to provide this amount of HCl, we use the following equation:

moles of solute = concentration × volume (in liters)

Rearranging for volume, we get:

volume = moles of solute / concentration

Plugging in the values, we get:

volume = 0.00507 mol / 0.6310 mol/L HCl = 0.00803 L = 8.03 mL

Therefore, we need 8.03 mL of 0.6310 M HCl to titrate 15.80 mL of 0.3210 M NaOH.