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

Answer:

$V_{base}=66.7mL$

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:

$V_{base}=\frac{M_{acid}V_{acid}}{M_{base}}$

Therefore, we plug in the given data to obtain:

$V_{base}=\frac{0.020M*50.0mL}{0.015M}\\\\ V_{base}=66.7mL$

Best regards!