A 25.0-mL sample of 0.150 M hypochlorous acid is titrated with a 0.150 M NaOH solution.What is the pH after 13.3 mL of base is added? The of hypochlorous acid is 3.0 × 10^-8.

Question

Quizplus · Accepted Answer

First, write the balanced chemical equation for the reaction between hypochlorous acid (HClO) and sodium hydroxide (NaOH): 
HClO + NaOH → NaClO + H2O

Next, determine the number of moles of HClO present in the initial solution: 
0.0250 L x 0.150 mol/L = 0.00375 mol HClO

Since NaOH and HClO react in a 1:1 ratio, the number of moles of NaOH added at the point where the pH is to be determined is also 0.00375 mol.

Now, calculate the amount of excess hydroxide ions (OH-) present in the solution after the addition of 13.3 mL (0.0133 L) of 0.150 M NaOH: 
0.0133 L x 0.150 mol/L = 0.001995 mol NaOH

Since NaOH and HClO react in a 1:1 ratio, the amount of excess OH- ions is also 0.001995 mol.

Therefore, the total amount of hydroxide ions in solution is: 
(0.00375 mol + 0.001995 mol) = 0.005745 mol

Finally, use the acid dissociation constant (Ka) and the initial concentration of HClO to calculate the pH:

Ka = [H+][ClO-]/[HClO] = 3.0 x 10^-8
[HClO] = 0.150 mol/L x 0.0250 L = 0.00375 mol
[OH-] = 0.005745 mol
[H+] = Ka*[HClO]/[OH-] = (3.0 x 10^-8)*(0.00375 mol)/(0.005745 mol) = 1.96 x 10^-8 M
pH = -log[H+] = 7.58

Therefore, the answer is D) 7.58.

A 250-ML Sample of 0