pH of solution after the addition of 23.9mL of 0.331 M HCl

Question

The p𝐾a of hypochlorous acid is 7.530. A 55.0 mL solution of 0.137 M sodium hypochlorite (NaOCl) is titrated with 0.331 M HCl. Calculate the pH of the solution after the addition of 23.9 mL of 0.331 M HCl.

23.9 mL - 0.0239 L

55.0 mL - 0.055 L

pH=pKa + log [ conjugated base/ weak acid]

7.530 + log [ 55.0* 0.137/ 23.9* 0.331]

7.530 + log [7.535 / 7.9109]

7.530 + log [ 0.952]

7.530 + (-0.021)= 7.509

pH= 7.509 ** is identified as incorrect, where did I go wrong??

Answer 1

pH = 2.32

This makes sense because a strong acid is the titrant and the other reagent is completely used up, passing the "equivalence point" in a titration curve and significantly dropping pH.

Edit: Initial answer incorrect, but I found the calculation error and edited the response. Let me know if this helps!

The equation is indeed 1:1:1:1 with NaClO + HCl ⇔ HClO + NaCl.

You used the formula of of pH=pKa + log([base]/[acid]), and you found the correct initial mole amounts of 0.007535 moles NaClO and 0.0079109 moles HCl.

However, the ([base]/[acid]) = [NaClO]/[HClO] . The issue here is that as [HClO] is being produced and increasing, [NaClO] is being used and decreasing. All of NaClO is used up, so the formula you used cannot work for this problem as there is leftover HCl without any NaClO. Instead, we can find [H⁺] and calculate the pH with pH=-log[H⁺].

Reacting fully, we can use subtraction to find the remaining HCl moles:

0.0079109 mol HCl - 0.007535 mol NaClO = 0.000374 mol HCl

We also need a total volume, so we add the given 55 mL + 23.9 mL = 78.9 mL of solution

Now we find [H⁺] by dividing [H⁺]= 0.000374 mol HCl / 0.0789 L solution = 0.004740177

pH= -log[H⁺]=-log(0.004740177)=2.3242= 2.32 (3 significant figures; assuming the produced moles of HClO don't have a significant contribution to [H⁺] given that its pKa is already accounted for)