Q19.37P - Chemistry: Molecular Nature Of Matter And Change

Question

An industrial chemist studying bleaching and sterilizing prepares several hypochlorite buffers. Find the pH of (a) $\mathbf{\text{0.100 M HClO}}$ and $\mathbf{\text{0.100 M NaClO}}$; (b) $\mathbf{\text{0.100 M HClO}}$ and $\mathbf{\text{0.150 M NaClO}}$; (c) $\mathbf{\text{0.150 M HClO}}$ and $\mathbf{\text{0.100 M NaClO}}$; (d) $\mathbf{\text{1.0 L}}$ of the solution in part (a) after $\mathbf{\text{0.0050}}$ mol of $\mathbf{\text{NaOH}}$ has been added.?

Step-by-Step Solution

Answer

1. The value of $\text{pH = 7.538}$.
2. The value of $\text{pH = 7.714}$.
3. The value of $\text{pH = 7.362}$.
4. The value of $\text{pH = 7.581}$.

Step 1: Define buffer

A buffer is a substance that can survive pH fluctuations caused by the addition of acidic or basic substances.It can neutralize little amounts of additional acid or base, allowing the pH of the solution to remain relatively constant.

Step 2: Explanation

(a)

First and foremost, the conjugated acid-base pair ${\text{HClO/ClO}}^{\text{ - }}$ must be mentioned,

$\text{NaClO}\to {\text{Na}}^{\text{ + }}{\text{ + ClO}}^{\text{ - }}$

which shows the concentration of ${\text{ClO}}^{\text{ - }}$ is the same as the concentration of $\text{NaClO}$.

To answer this problem, we only need to use the Henderson-Hasselbalch equation:

$$\begin{gathered} \mathrm{pH}=\mathrm{pKa}+\log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ \mathrm{Ka}\left(\mathrm{HClO}\right)=2.9\times {10}^{-8} \\ \mathrm{pKa}=-\mathrm{logKa} \\ =7.538 \\ \mathrm{pH}=7.538+\log \frac{0.1 \mathrm{M}}{0.1 \mathrm{M}} \\ =7.538 \end{gathered}$$

Therefore, $\text{pH = 7.538}$.

Step 3: Explanation

(b) 

The principle is the same as it was in the previous case:

$$\begin{gathered} \mathrm{pH}=\mathrm{pKa}+\log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ \mathrm{pH}=7.538+\log \frac{0.15 \mathrm{M}}{0.1 \mathrm{M}} \\ =7.714 \end{gathered}$$

Therefore, $\text{pH = 7.714}$.

Step 4: Explanation

(c)

The principle is the same as it was in the previous case:

$$\begin{gathered} \mathrm{pH}=\mathrm{pKa}+\log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ \mathrm{pH}=7.538+\log \frac{0.1 \mathrm{M}}{0.15 \mathrm{M}} \\ =7.362 \end{gathered}$$

Therefore, $\text{pH = 7.362}$.

Step 5: Explanation

(d)

• $1.0 \mathrm{L}$of the solution in section (a) after adding $0.0050$ mol of $\text{NaOH}$ ${\text{HClO(aq) + OH}}^{\text{ - }}{\text{ > > H}}_{\text{2}}{\text{O(l) + ClO}}^{\text{ - }}$

• As $\text{NaOH}$ is the limiting reagent and all of it has reacted, the initial number of moles ${\mathrm{OH}}^{-}$ of is $0.005$ mol, but at equilibrium it is 0 mol.
• The initial number of moles $\text{HClO}$ of is $\text{0.1}$ mol ($\mathrm{n}=\mathrm{c}\times \mathrm{V}$), however it $0.1 \mathrm{mol}-0.005 \mathrm{mol}=0.095 \mathrm{mol}$ is in the equilibrium state.
• The initial number of moles of ${\mathrm{ClO}}^{-}$ is $0.1$mol, however it is  $0.1 \mathrm{mol}+0.005 \mathrm{mol}=0.105 \mathrm{mol}$ in equilibrium.
• In the equilibrium state, the M values of $\text{HClO}$ and ${\text{ClO}}^{\text{ - }}$ are $\text{0.095 M}$ and $\text{0.105 M}$, respectively ( $\mathrm{c}=\mathrm{n}/\mathrm{V}$).

          Now we'll use the Henderson-Hasselbalch formula:

           $$\begin{gathered} \mathrm{pH}=\mathrm{pKa}+\log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ \mathrm{pH}=7.538+\log \frac{0.105 \mathrm{M}}{0.095 \mathrm{M}} \\ =7.581 \end{gathered}$$

         

          Therefore, $\text{pH = 7.581}$.