How many milliliters of 0.120 M NaOH are required to titrate 50.0 mL of 0.0998 M hypochlorous acid to the equivalence point? The K_a of hypochlorous acid is 3.0 × 10^-8.

Question

Quizplus · Accepted Answer

Step 1:
Write the balanced chemical equation for the reaction between NaOH and hypochlorous acid (HOCl):
HOCl + NaOH → NaOCl + H2O

Step 2:
Determine the number of moles of HOCl present in the 50.0 mL of 0.0998 M solution:
0.0500 L x 0.0998 mol/L = 0.00499 mol

Step 3:
Use the balanced equation to determine the number of moles of NaOH required to react with the HOCl:
1 mol HOCl reacts with 1 mol NaOH
Therefore, 0.00499 mol HOCl requires 0.00499 mol NaOH

Step 4:
Calculate the volume of 0.120 M NaOH required to provide 0.00499 mol:

0.00499 mol NaOH x (1 L/0.120 mol) x (1000 mL/1 L) = 41.6 mL

Therefore, the answer is (C) 41.6.

How Many Milliliters of 0