Question:
The pH of 1 L of HCl solution is 1.0. What is the volume (in L) of water to be added to this solution to increase its pH to 2.0?
The pH of 1 L of HCl solution is 1.0. What is the volume (in L) of water to be added to this solution to increase its pH to 2.0?
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A quick rule of thumb: to increase the pH of a strong acid by 1 unit, the solution must be diluted to 10 times its original volume.
Therefore, the final volume is 10 L, which means we must add $(10 - 1) = 9$ L of water.
Therefore, the final volume is 10 L, which means we must add $(10 - 1) = 9$ L of water.
Updated On: Jul 22, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The question is about dilute acid chemistry.
We need to determine the volume of water required to dilute a hydrochloric acid (HCl) solution to increase its pH from 1.0 to 2.0.
Step 2: Key Formula or Approach:
First, we find the hydrogen ion concentrations $[\text{H}^+]$ from the pH values:
\[ [\text{H}^+] = 10^{-\text{pH}} \] Then, we use the dilution law to calculate the final volume:
\[ M_1 V_1 = M_2 V_2 \] Finally, the volume of water to be added is:
\[ V_{\text{added}} = V_2 - V_1 \]
Step 3: Detailed Explanation:
• Let us calculate the initial concentration ($M_1$) of the solution:
Initial pH = 1.0 $\implies M_1 = [\text{H}^+]_1 = 10^{-1.0} = 0.1\text{ M}$.
Initial volume ($V_1$) = $1\text{ L}$.
• Let us calculate the final concentration ($M_2$) required:
Final pH = 2.0 $\implies M_2 = [\text{H}^+]_2 = 10^{-2.0} = 0.01\text{ M}$.
• Apply the dilution formula:
\[ M_1 V_1 = M_2 V_2 \] \[ 0.1 \times 1 = 0.01 \times V_2 \] \[ V_2 = \frac{0.1}{0.01} = 10\text{ L} \]
• Calculate the volume of water that must be added:
\[ V_{\text{added}} = V_2 - V_1 = 10\text{ L} - 1\text{ L} = 9\text{ L} \]
Step 4: Final Answer:
The volume of water to be added is 9 L.
The question is about dilute acid chemistry.
We need to determine the volume of water required to dilute a hydrochloric acid (HCl) solution to increase its pH from 1.0 to 2.0.
Step 2: Key Formula or Approach:
First, we find the hydrogen ion concentrations $[\text{H}^+]$ from the pH values:
\[ [\text{H}^+] = 10^{-\text{pH}} \] Then, we use the dilution law to calculate the final volume:
\[ M_1 V_1 = M_2 V_2 \] Finally, the volume of water to be added is:
\[ V_{\text{added}} = V_2 - V_1 \]
Step 3: Detailed Explanation:
• Let us calculate the initial concentration ($M_1$) of the solution:
Initial pH = 1.0 $\implies M_1 = [\text{H}^+]_1 = 10^{-1.0} = 0.1\text{ M}$.
Initial volume ($V_1$) = $1\text{ L}$.
• Let us calculate the final concentration ($M_2$) required:
Final pH = 2.0 $\implies M_2 = [\text{H}^+]_2 = 10^{-2.0} = 0.01\text{ M}$.
• Apply the dilution formula:
\[ M_1 V_1 = M_2 V_2 \] \[ 0.1 \times 1 = 0.01 \times V_2 \] \[ V_2 = \frac{0.1}{0.01} = 10\text{ L} \]
• Calculate the volume of water that must be added:
\[ V_{\text{added}} = V_2 - V_1 = 10\text{ L} - 1\text{ L} = 9\text{ L} \]
Step 4: Final Answer:
The volume of water to be added is 9 L.
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