We are asked to **calculate the pH change of the solution. **

We can see that the solution has **HClO and NaClO** which is the salt for the conjugate base of HClO.

** HClO + H _{2}O → ClO^{-} + H_{3}O^{+}**

This means we'll have a buffer.

We use the formula below to calculate the pH of a buffer Henderson-Hasselbalch equation:

$\overline{){\mathbf{pH}}{\mathbf{=}}{{\mathbf{pK}}}_{{\mathbf{a}}}{\mathbf{+}}{\mathbf{log}}\left(\frac{\mathbf{conjugate}\mathbf{}\mathbf{base}}{\mathbf{weak}\mathbf{}\mathbf{acid}}\right)}$

**We go through the following steps to solve the problem: **

**Step 1. **Calculate the moles of ClO^{-}

**Step 2. **Calculate the moles HClO

**Step 3. **Calculate the pK_{a}

**Step 4.** Calculate the pH of the solution with NaClO added

**Step 5. **Calculate the pH change

**Step 1.** Calculate the moles of NaClO

We will use the molar mass or the formula weight (FW):

$\mathbf{Moles}\mathbf{}{\mathbf{ClO}}^{\mathbf{-}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{68}\mathbf{}\overline{)\mathbf{g}\overline{)\mathbf{}\mathbf{NaClO}}}\mathbf{}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{NaClO}}}{\mathbf{74}\mathbf{.}\mathbf{44}\mathbf{}\overline{)\mathbf{g}\overline{)\mathbf{}\mathbf{NaClO}}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{ClO}}^{\mathbf{-}}}{\mathbf{1}\overline{)\mathbf{}\mathbf{mol}\mathbf{}\mathbf{NaClO}}}$

**Moles ClO ^{-} = 0.0226 mol ClO**

The pH of 0.50 M HClO is 3.91. Calculate the change in pH when 1.68 g of NaClO (FW = 74.44 g/mol) is added to 31 mL of 0.50 M HClO (FW = 52.46 g/mol). Ignore any changes in volume. The K_{a} value for HClO is 3.0 x 10^{-8} . Enter your answer with two decimal places.

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What scientific concept do you need to know in order to solve this problem?

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