A 25.0 mL sample of 0.723 M HClO4 is titrated with a KOH solution. The H3O+ concentration after the addition of of KOH is
To determine the H3O+ concentration after the addition of KOH, we need to first calculate the moles of HClO4 in the 25.0 mL sample.
Step 1: Calculate the moles of HClO4:
Moles = Molarity * Volume (L)
= 0.723 M * 0.0250 L
= 0.018075 moles
Step 2: Determine the stoichiometry of the reaction between HClO4 and KOH. The balanced equation is:
HClO4 + KOH → KClO4 + H2O
From the balanced equation, we can see that 1 mole of HClO4 reacts with 1 mole of KOH to produce 1 mole of H2O.
Step 3: Calculate the moles of H3O+ formed after the reaction:
Since we have equal stoichiometry for HClO4 and H3O+ in the balanced equation, the moles of H3O+ formed will also be equal to the moles of HClO4.
Moles of H3O+ = 0.018075 moles
Step 4: Calculate the concentration of H3O+ in the final solution:
Final Volume = Initial Volume + Volume of KOH added
= 25.0 mL + (Volume of KOH added)
Since the volume of KOH added is not given in the question, we cannot determine the final volume or concentration of H3O+ without this information.