Respuesta :
Answer:
For a: The volume of NaOH used in the titration is 180 mL
For b: The volume of NaOH used in the titration is 200 mL
Explanation:
To calculate the volume of base, 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
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base
- For a:
We are given:
[tex]n_1=1\\M_1=0.600M\\V_1=30.00mL\\n_2=1\\M_2=0.100M\\V_2=?mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.600\times 30.00=1\times 0.100\times V_2\\\\V_2=\frac{1\times 0.600\times 30}{1\times 0.100}=180mL[/tex]
Hence, the volume of NaOH used in the titration is 180 mL
- For b:
We are given:
[tex]n_1=1\\M_1=0.400M\\V_1=50.00mL\\n_2=1\\M_2=0.100M\\V_2=?mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.400\times 50.00=1\times 0.100\times V_2\\\\V_2=\frac{1\times 0.400\times 50}{1\times 0.100}=200mL[/tex]
Hence, the volume of NaOH used in the titration is 200 mL
