Question

What volume in mL of your HCl solution did you use for each of the three acceptable trial samples for the standardization? Your volume of HCl should be approximately 5 mL. Report your volume to 4 significant figures.

Answer 1

The question is incomplete, here is the complete question:

What volume in mL of your HCl solution did you use for each of the three acceptable trial samples for the standardization?

Trials Molarity of HCl Volume of NaOH

1 0.1 M 24.80 mL

2 0.2 M 19.20 mL

3 0.01 M 18.00 mL

The average molarity of NaOH is 0.0755 M

Answer:

For Trial 1: The volume of HCl required is 18.72 mL

For Trial 2: The volume of HCl required is 7.25 mL

For Trial 3: The volume of HCl required is 135.9 mL

Step-by-step explanation:

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

`n_1M_1V_1=n_2M_2V_2`

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=0.1M\\V_1=?mL\\n_2=1\\M_2=0.0755M\\V_2=24.80mL`

Putting values in above equation, we get:

`1* 0.1* V_1=1* 0.0755* 24.80\\\\V_1=(1* 0.0755* 24.80)/(1* 0.1)=18.72mL`

Hence, the volume of HCl required is 18.72 mL

• For Trial 2:

We are given:

`n_1=1\\M_1=0.2M\\V_1=?mL\\n_2=1\\M_2=0.0755M\\V_2=19.20mL`

Putting values in above equation, we get:

`1* 0.2* V_1=1* 0.0755* 19.20\\\\V_1=(1* 0.0755* 19.20)/(1* 0.2)=7.25mL`

Hence, the volume of HCl required is 7.25 mL

• For Trial 3:

We are given:

`n_1=1\\M_1=0.01M\\V_1=?mL\\n_2=1\\M_2=0.0755M\\V_2=18.00mL`

Putting values in above equation, we get:

`1* 0.01* V_1=1* 0.0755* 18.00\\\\V_1=(1* 0.0755* 18.00)/(1* 0.01)=135.9mL`

Hence, the volume of HCl required is 135.9 mL