The balanced chemical equation for the reaction between HClO and NaOH is:
HClO(aq) + NaOH(aq) → NaClO(aq) + H2O(l)
From the equation, we can see that the stoichiometry of the reaction is 1:1, which means that 1 mole of HClO reacts with 1 mole of NaOH.
Before any NaOH is added, we have 0.100 M HClO in a 250.0 mL solution. To determine the number of moles of HClO in the solution, we use the equation:
Molarity × Volume = moles
Moles of HClO = Molarity × Volume = 0.100 mol/L × 0.250 L = 0.0250 mol
Since the stoichiometry of the reaction is 1:1, the number of moles of NaOH required to react with the HClO is also 0.0250 mol.
Now we can calculate the concentration of HClO after 50.0 mL of 0.200 M NaOH has been added. The number of moles of NaOH added is:
Molarity × Volume = moles
Moles of NaOH = Molarity × Volume = 0.200 mol/L × 0.0500 L = 0.0100 mol
Since the stoichiometry of the reaction is 1:1, the number of moles of HClO that remains after the reaction is:
Moles of HClO remaining = Moles of HClO - Moles of NaOH = 0.0250 mol - 0.0100 mol = 0.0150 mol
The volume of the solution after the NaOH is added is:
Volume = initial volume + volume of NaOH added = 0.250 L + 0.0500 L = 0.300 L
Therefore, the concentration of HClO after the NaOH is added is:
Concentration = Moles of HClO remaining / Volume of solution = 0.0150 mol / 0.300 L = 0.0500 M
To calculate the pH of the solution, we need to first determine the pKa of HClO. The pKa of HClO is 7.54. We can use the equation for the acid dissociation constant to calculate the concentration of H+:
Ka = [H+][ClO