Respuesta :
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
Answer:
For Trial 1: The molarity of HCl for trial 1 is 0.195 M
For Trial 2: The molarity of HCl for trial 2 is 0.191 M
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex] ......(1)
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is HCl
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
- For Trial 1:
We are given:
[tex]n_1=1\\M_1=?M\\V_1=10.00mL\\n_2=1\\M_2=0.15M\\V_2=13mL[/tex]
Putting values in equation 1, we get:
[tex]1\times M_1\times 10.00=1\times 0.15\times 13\\\\M_1=\frac{1\times 0.15\times 13}{1\times 10.00}=0.195M[/tex]
Hence, the molarity of HCl for trial 1 is 0.195 M
- For Trial 2:
We are given:
[tex]n_1=1\\M_1=?M\\V_1=10.00mL\\n_2=1\\M_2=0.15M\\V_2=12.75mL[/tex]
Putting values in equation 1, we get:
[tex]1\times M_1\times 10.00=1\times 0.15\times 12.75\\\\M_1=\frac{1\times 0.15\times 12.75}{1\times 10.00}=0.191M[/tex]
Hence, the molarity of HCl for trial 2 is 0.191 M
