91Ó°ÊÓ

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.

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

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:

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

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

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

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

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

• 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 role="math" localid="1663320150006" $\text{HClO}$of is $\text{0.1}$mol ($\mathrm{n}=\mathrm{c}×\mathrm{V}$), 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 isrole="math" localid="1663320389237" $0.1\mathrm{mol}+0.005\mathrm{mol}=0.105\mathrm{mol}$ 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 (role="math" localid="1663320540360" $\mathrm{c}=\mathrm{n}/\mathrm{V}$).

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, $\text{pH = 7.581}$.