Suppose a student adds 25.00 mL of 1.025 M HCl to a 1.50 g antacid tablet. The student boils and then titrates the resulting solution to the endpoint with 0.4969 M NaOH. The titration requires 21.3 mL NaOH to reach the endpoint. How many moles of HCl were neutralized by the NaOH?
To find the number of moles of HCl neutralized by NaOH, we can start by calculating the number of moles of NaOH used in the titration.
1. Firstly, we need to convert the volume of NaOH used to liters:
21.3 mL = 21.3 mL * (1 L / 1000 mL) = 0.0213 L
2. Next, we can calculate the number of moles of NaOH used using its molar concentration (0.4969 M) and the volume used:
moles of NaOH = molar concentration * volume
= 0.4969 M * 0.0213 L
Now, we have the number of moles of NaOH used in the titration. However, since NaOH and HCl react in a 1:1 ratio, the number of moles of HCl neutralized by NaOH is equal to the number of moles of NaOH used.
Therefore, the number of moles of HCl neutralized by NaOH is also 0.4969 M * 0.0213 L.
Please note that the concentration of HCl added to the solution (1.025 M) and the volume added (25.00 mL) are not relevant to finding the number of moles neutralized, as the titration endpoint is solely determined by the reaction between NaOH and HCl.