Question

$10^{-6} \text{ M } \text{NaOH}$ is diluted 100 times. The $\text{pH}$ of the diluted base is:

• A. between 7 and 8
• B. between 5 and 6
• C. between 6 and 7
• D. between 10 and 11

Hint:

When the acid or base concentration falls below $10^{-6} \text{ M}$, the self-ionization of water must be accounted for. The final $\text{pH}$ of any aqueous solution at $25^{\circ}\text{C}$ cannot exceed the range $[0, 14]$ and cannot be far from 7 for ultra-dilute solutions.

Answer 1

The Correct Option is A

The initial concentration of $\text{OH}^-$ is $[\text{OH}^-]_{\text{initial}} = 10^{-6} \text{ M}$.
After 100 times dilution, the $[\text{OH}^-]$ from the base is:
$[\text{OH}^-]_{\text{base}} = \frac{10^{-6}}{100} = 10^{-8} \text{ M}$.
Since this concentration is comparable to $[\text{OH}^-]$ from water auto-ionization ($10^{-7} \text{ M}$), the total $[\text{OH}^-]$ must be considered.
Total $[\text{OH}^-]$ will be slightly greater than $10^{-7} \text{ M}$.
If $[\text{OH}^-] = 10^{-7} \text{ M}$, $\text{pOH} = 7$ and $\text{pH} = 7$.
Since the total $[\text{OH}^-]$ is greater than $10^{-7} \text{ M}$ (due to the added base), the $\text{pOH}$ will be slightly less than 7.
Therefore, the $\text{pH} = 14 - \text{pOH}$ will be slightly greater than 7.
The $\text{pH}$ of the diluted solution is between 7 and 8.