Question

calculate the pH for each of the following cases in the titration of 50.0 ml of .180 M HClO(aq) with .180 M KOH (aq)

25 ml of KOH
35 ml of KOH
50 ml of KOH
60 ml of KOH

Answer 1

the reaction is HClO + KOH ------> KClO + H2O
find the moles of each by multiplying the molarity with volume in liters.
HClO ----> 0.050 X 0.18 = 9 x 10^-3
KOH ---> 0.025 X 0.18 = 4.5 x 10^-3
its mean after complete reaction excess moles of HClO are remaining, which is acid.
now find how many excess moles of acid are remaining, that will be,
9 x 10^-3 - 4.5 x 10^-3 = 4.5 x 10^-3 moles of HClO
now use the equation,
HClO ------> H+ + ClO-
4.5 x 10^-3 moles 4.5 x 10^-3 moles of H+

convert these moles into concentration by dividing them with total volume,
i.e 0.050 + 0.025 = 0.075
concentration of H+ = 4.5 x 10^-3 / 0.075 = 0.06 M

pH = -log [H+]
pH = - log 0.06
pH = 1.22

Like this u can calculate rest of questions.