Question

What volume of 0.235 M H2SO4 is needed to titrate 40.0 mL of 0.0500 M Na2CO M NaCO3 solution?

Answer 1

Answer: The volume of 0.235 M
`H_2SO_4` needed to titrate 40.0 mL of 0.0500 M
`Na_2CO_3` is 8.51 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
`H_2SO_4`

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is
`Na_2CO_3`

We are given:

`n_1=2\\M_1=0.235M\\V_1=?mL\\n_2=2\\M_2=0.0500M\\V_2=40.0mL`

Putting values in above equation, we get:

`2* 0.235* V_1=2* 0.0500* 40.0\\\\V_1=8.51mL`

Thus volume of 0.235 M
`H_2SO_4` needed to titrate 40.0 mL of 0.0500 M
`Na_2CO_3` is 8.51 ml