The question is incomplete, here is the complete question:
For each of your 2 ACCEPTABLE trials enter the values you calculated for the molarity of your HCl solution in the same sequence corresponding to the endpoint volumes.
Endpoint (in mL):
Trial 1: 13 mL
Trial 2: 12.75 mL
Other info:
10.00 mL of HCl solution in beaker for each trial. 0.15 M NaOH solution in burrett for titration
<u>Answer:</u>
<u>For Trial 1:</u> The molarity of HCl for trial 1 is 0.195 M
<u>For Trial 2:</u> The molarity of HCl for trial 2 is 0.191 M
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
......(1)
where,
are the n-factor, molarity and volume of acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in equation 1, we get:
Hence, the molarity of HCl for trial 1 is 0.195 M
We are given:
Putting values in equation 1, we get:
Hence, the molarity of HCl for trial 2 is 0.191 M