🤓 Based on our data, we think this question is relevant for Professor Dixon's class at UCF.
The reaction between HClO and KOH is given by this balanced equation:
HClO (aq) + KOH (aq) à KClO (aq) + H2O (aq)
Since HClO has a Ka value of 3.5x10-8 which is less than 1, it is a weak acid. Thus, this problem tackles a weak acid – strong base titration.
From the balanced equation, a mole-to mole comparison shows that 1 mole of HClO is equal to 1 mole of KOH.
This means that at every stage of the titration, the number of moles of the base reacting to the acid can be calculated.
The pH of the solution can then be determined using the excess acid or base that is not consumed in the reaction.
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.250 M HClO(aq) with 0.250 M KOH(aq). The ionization constant for HClO can be found here.
(a) before addition of any KOH
(b) after addition of 25.0 mL of KOH
(c) after addition of 30.0 mL of KOH
(d) after addition of 50.0 mL of KOH
(e) after addition of 60.0 mL of KOH