Question

For the titration of 50.0 mL of .020M HI with 0.015 M of NaOH, graph pH versus milliliters of base added from 0-100 mL. How many milliliters of NaOH are added at the equivalence point?

Answer 1

`V_(base)=66.7mL`

Step-by-step explanation:

Hello there!

In this case, since the neutralization reaction between HI and NaOH is:

`NaOH+HI\rightarrow NaI+H_2O`

Thus, as there is a 1:1 mole ratio of base to acid, it is possible to use the following mole equivalence:

`n_(base)=n_(acid)\\\\M_(base)V_(base)=M_(acid)V_(acid)`

Thus, by solving for the volume of base, we obtain:

`V_(base)=(M_(acid)V_(acid))/(M_(base))`

Therefore, we plug in the given data to obtain:

`V_(base)=(0.020M*50.0mL)/(0.015M)\\\\ V_(base)=66.7mL`

Best regards!