a 30.0 ml sample of 0.75 M hypociouous acid, hocl, is titrated with 1.25 M KOH. what is the ph of the solution after 7.0 ml of KOH is added
[HOCl] = molarity x volume in Litres = 0.75 M x 0.03 L = 0.0225 mol
[KOH] = molarity x volume in Litres = 1.25 M x 0.007 L = 0.00875 mol
HOCl + KOH -----------------> KOCl + H2O
0.0225 mol 0.00875 mol
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0.0225 -0.00875 0 0.00875 mol
= 0.01375 mol
Hence,
[HOCl] = 0.01375 mol
[KOCl] = 0.00875 mol
We know that pKa of HOCl = 7.53
pH = pKa + log [salt] / [acid]
= pKa + log [KOCl] / [HOCl]
= 7.53 + log (0.00875/0.01375)
= 7.33
pH = 7.33
Therefore,
pH of the given solution = 7.33
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