We are being asked to determine the **pH of the HClO solution**.

**(a) before addition of any KOH**** **

The dissociation of HClO in water:

**HClO + H _{2}O **

$\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{}\mathbf{M}\mathbf{}\overline{)\mathbf{HClO}}\left(\frac{1\overline{)\mathrm{mol}}{H}_{3}{O}^{+}}{1\overline{)\mathrm{mol}\mathrm{HClO}}}\right)\mathbf{=}$**0.250 M H _{3}O**

$\overline{){\mathbf{pH}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{-}}{\mathbf{log}}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}}\phantom{\rule{0ex}{0ex}}\mathbf{pH}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{log}\mathbf{[}\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{}\mathbf{M}\mathbf{]}$

**pH = 0.60**

**(b) after addition of 25.0 mL of KOH**

**We will calculate the pH of the HClO solution using the following steps:**

*Step 1**. Calculate the initial amounts of HClO and KOH in moles before the reaction happens.*

**Step 2.** Write the **chemical equation** for the reaction between HClO and KOH.

**Step 3.** Construct an **ICF Chart**.

**Step 4.** Calculate pH.

**Step 1**. Calculate the **initial amounts** of HClO and NaOH in moles before the reaction happens.

**molarity (volume) → moles**

Recall:

$\overline{){\mathbf{Molarity}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{Liters}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}}$

**convert volumes from mL to L → 1 mL = 10 ^{-3} L*

**•**** 50.0 mL of 0.250 M HClO(aq)**

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