Question

For each of your NCount ACCEPTABLE trials enter the values you calculated for the molarity of your HCl solution in the same sequence corresponding to the equivalence point volumes. You should report the molarities to 4 significant figures, e.g. 0.2314 M.

Answer 1

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

Step-by-step explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

`n_1M_1V_1=n_2M_2V_2` ......(1)

where,

`n_1,M_1\text{ and }V_1` are the n-factor, molarity and volume of acid which is HCl

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is NaOH.

• For Trial 1:

We are given:

`n_1=1\\M_1=?M\\V_1=10.00mL\\n_2=1\\M_2=0.15M\\V_2=13mL`

Putting values in equation 1, we get:

`1* M_1* 10.00=1* 0.15* 13\\\\M_1=(1* 0.15* 13)/(1* 10.00)=0.195M`

Hence, the molarity of HCl for trial 1 is 0.195 M

• For Trial 2:

We are given:

`n_1=1\\M_1=?M\\V_1=10.00mL\\n_2=1\\M_2=0.15M\\V_2=12.75mL`

Putting values in equation 1, we get:

`1* M_1* 10.00=1* 0.15* 12.75\\\\M_1=(1* 0.15* 12.75)/(1* 10.00)=0.191M`

Hence, the molarity of HCl for trial 2 is 0.191 M