Consider a titration of 50.00 mL of 0.1005 M HCl with 0.1012 M NaOH.
To find the result of this titration, we need to determine the quantity of the HCl that reacts with the NaOH. This can be done by using the concept of stoichiometry and the equation that represents the reaction between HCl and NaOH:
HCl + NaOH -> NaCl + H2O
First, we need to find the number of moles of HCl. We can use the formula:
moles of solute = molarity × volume (in liters)
moles of HCl = 0.1005 M × 0.05000 L
= 0.005025 moles
According to the balanced chemical equation, the stoichiometric ratio between HCl and NaOH is 1:1. Therefore, 0.005025 moles of NaOH is required to react completely with the HCl.
Next, we need to determine the volume of NaOH required to react with the HCl. We can use the formula:
volume = moles of solute / molarity
volume of NaOH = 0.005025 moles / 0.1012 M
= 0.04965 L
Since the given volume of HCl is 50.00 mL, we can see that NaOH is in excess since 0.04965 L is less than 50.00 mL. This means that all the HCl will react with the NaOH, leaving some of the NaOH unreacted.
To find the remaining excess NaOH, we can subtract the volume of NaOH used from the initial volume:
excess NaOH = 0.05000 L - 0.04965 L
= 0.00035 L
In terms of milliliters, this is:
excess NaOH = 0.00035 L × 1000 mL/L
= 0.35 mL
Thus, there will be an excess of 0.35 mL of NaOH remaining after the titration.