Calculate the pH during the titration of 10.00 mL of 0.400 M hypochlorous acid with 0.500 M NaOH.
The Ka for HOCl is 3.0 x 10-8 M.
What is the pH when half the acid has been neutralized?
'I got 7.82 by using the HH formula but its wrong, help.
To calculate the pH when half the acid has been neutralized, we need to determine the concentration of hypochlorous acid (HOCl) and sodium hypochlorite (NaOCl) at the halfway point of the titration.
In this titration, HOCl (acid) reacts with NaOH (base) to form NaOCl (conjugate base) and water. The balanced equation for this reaction is:
HOCl + NaOH → NaOCl + H2O
Now, let's calculate the concentration of HOCl and NaOCl at the halfway point:
1. Initially, we have 10.00 mL of 0.400 M HOCl. This gives us:
Initial moles of HOCl = 0.01000 L × 0.400 M = 0.00400 mol HOCl
2. At the halfway point, half of the HOCl will be neutralized. This results in the formation of an equal number of moles of NaOCl:
Moles of NaOCl = Moles of HOCl neutralized = 0.00400 mol HOCl / 2 = 0.00200 mol NaOCl
Now, let's calculate the concentration of NaOCl at the halfway point:
Final volume = 10.00 mL HOCl + 10.00 mL NaOH = 0.02000 L
Concentration of NaOCl = Moles of NaOCl / Final volume = 0.00200 mol / 0.02000 L = 0.100 M NaOCl
Since NaOCl is the conjugate base of HOCl, it hydrolyzes in water to produce hydroxide ions (OH-).
Now, using the concentration of NaOCl at the halfway point, we can calculate the concentration of OH- ions:
OH- concentration = concentration of NaOCl = 0.100 M
Remember, in a neutralization reaction, the acid reacts with the base to form water. So, the concentration of hydroxide ions (OH-) at the halfway point represents the concentration of hydroxide ions resulting from the complete neutralization of the acid.
Finally, to calculate the pOH at the halfway point:
pOH = -log10(OH- concentration) = -log10(0.100) = 1.00
To find the pH, we use the relationship:
pH + pOH = 14
pH = 14 - pOH = 14 - 1.00 = 13.00
Therefore, the pH at the halfway point of the titration is approximately 13.00, not 7.82.