At 25°C, enough distilled water is added to a 30.0 mL sample of HNO3(aq) with a pH of 4.20 so that t — Acids and Bases Chemistry Question

STEPS:

1. Analyze the initial conditions and target change in pH:
The initial solution of strong acid $\text{HNO}_3(aq)$ has a pH of $4.20$ and a volume of $30.0\text{ mL}$. Distilled water is added to dilute the solution until the final pH reaches $5.20$. The overall change in pH is:
$$\Delta\text{pH} = 5.20 - 4.20 = \mathbf{1.00\text{ unit}}$$

2. Apply the logarithmic definition of pH to determine the dilution factor:
The pH scale is logarithmic, where $\text{pH} = -\log[\text{H}^+]$. Because of this logarithmic relationship, an increase in pH of exactly $1.00$ unit requires the hydronium/hydrogen ion concentration ($[\text{H}^+]$) of the acid to decrease by a factor of $10$.

3. Calculate the final total volume of the solution ($V_f$):
Since $\text{HNO}_3$ is a strong acid that dissociates completely in water, decreasing its $[\text{H}^+]$ concentration by a factor of $10$ requires diluting the entire solution to $10$ times its original volume:
$$V_f = V_i \times 10$$
$$V_f = 30.0\text{ mL} \times 10 = \mathbf{300.\text{ mL}} \quad$$

4. Calculate the volume of distilled water that must be added ($V_{\text{added}}$):
The final total volume ($V_f$) is equal to the initial sample volume ($V_i$) plus the volume of distilled water added ($V_{\text{added}}$):
$$V_f = V_i + V_{\text{added}}$$
$$300.\text{ mL} = 30.0\text{ mL} + V_{\text{added}} \quad$$
$$V_{\text{added}} = 300.\text{ mL} - 30.0\text{ mL} = \mathbf{270.\text{ mL}} \quad$$
This calculation shows that adding approximately $270.\text{ mL}$ of distilled water achieves the required dilution, which directly corresponds to Option C.

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WHY_OTHERS_WRONG:

• Option A is incorrect (30.0 mL): Adding only $30.0\text{ mL}$ of water would double the solution's total volume to $60.0\text{ mL}$. Doubling the volume only cuts the $[\text{H}^+]$ concentration in half, which changes the pH by $\log(2) \approx 0.30$ units (resulting in a pH of $4.50$), well short of the target $5.20$.
• Option B is incorrect (60.0 mL): Adding $60.0\text{ mL}$ of water results in a final volume of $90.0\text{ mL}$. This $3\text{-fold}$ dilution decreases the concentration of $[\text{H}^+]$ by a factor of $3$, which only increases the pH by $\log(3) \approx 0.48$ units (resulting in a pH of $4.68$).
• Option D is incorrect (300. mL): $300.\text{ mL}$ represents the final *total* volume of the diluted solution, not the volume of water to be added. If a student mistakenly adds $300.\text{ mL}$ of water to the original $30.0\text{ mL}$ sample, the final volume becomes $330.\text{ mL}$, resulting in an over-dilution.