[wizrak]

5.0 mL of 0.100 mol L–1 NaOH(aq) is added to a mixture of 10.0 mL of 0.200 mol L–1
HOBr(aq) and 10.0 mL of 0.200 mol L–1 CH3COOH(aq).
Consider the relative strengths of the acids and calculate the pH of the resulting mixture.

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Mark scheme
n(NaOH) = 0.0050 × 0.100 = 5.0 × 10–4 mol = n(CH3COO–)
n(CH3COOH) left = 2.0 × 10–3 – 5.0 × 10–4 = 1.5 × 10–3 mol
reconfiguration of K_A to find [H+] and thus pH.

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My Question
1. why dont you take into account of H⁺ from dissociation of HOBr

*ps. If the answer is "It gets too complicated" please write out the real correct answer with the complicated working outs. Thanks

[Borek]

What are dissociation constants for both acids? (Ka, pKa, whatever form you have)

[wizrak]

pKa of ethanoic acid is 4.8 (I read this from a titration curve at 1/2 equiv point)
pKa of HOBr is 8.69

[Borek]

So HOBr is about 10000 times weaker. That means in typical situation (when they are both present in similar concentration, and not very diluted) its effect on the final pH will be about 10000 times lower than the effect of the acetic acid. Do you see now why it can be ignored?

[wizrak]

Thank you Borek
I understand it now