Question

a 30.0 ml sample of 0.75 M hypociouous acid, hocl, is titrated with 1.25 M KOH....

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

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Answer #1

[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   

-------------------------------------------------------------------------------------------------------------

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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