Respuesta :
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
Explanation:
To calculate the volume of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
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=0.1M\\V_1=?mL\\n_2=1\\M_2=0.0755M\\V_2=24.80mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.1\times V_1=1\times 0.0755\times 24.80\\\\V_1=\frac{1\times 0.0755\times 24.80}{1\times 0.1}=18.72mL[/tex]
Hence, the volume of HCl required is 18.72 mL
- For Trial 2:
We are given:
[tex]n_1=1\\M_1=0.2M\\V_1=?mL\\n_2=1\\M_2=0.0755M\\V_2=19.20mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.2\times V_1=1\times 0.0755\times 19.20\\\\V_1=\frac{1\times 0.0755\times 19.20}{1\times 0.2}=7.25mL[/tex]
Hence, the volume of HCl required is 7.25 mL
- For Trial 3:
We are given:
[tex]n_1=1\\M_1=0.01M\\V_1=?mL\\n_2=1\\M_2=0.0755M\\V_2=18.00mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.01\times V_1=1\times 0.0755\times 18.00\\\\V_1=\frac{1\times 0.0755\times 18.00}{1\times 0.01}=135.9mL[/tex]
Hence, the volume of HCl required is 135.9 mL
