You have 100.0 mL of a solution of hydrochloric acid that has a pH of 3.00. You add 100.0 mL of water to this solution. What is the resulting pH of the solution?
The Correct Answer and Explanation is :
To determine the resulting pH after diluting a hydrochloric acid (HCl) solution with water, we need to start by understanding the concentration of the original solution.
Step 1: Calculate the concentration of HCl in the original solution.
The pH of a solution is related to the concentration of hydrogen ions ([H^+]) through the equation:
[
\text{pH} = -\log[H^+]
]
Given that the pH is 3.00, we can calculate the concentration of hydrogen ions:
[
[H^+] = 10^{-\text{pH}} = 10^{-3} \, \text{M} = 0.001 \, \text{M}
]
Since HCl is a strong acid, it dissociates completely in solution. Therefore, the concentration of HCl in the original solution is also 0.001 M.
Step 2: Calculate the total volume after dilution.
When you add 100.0 mL of water to 100.0 mL of the HCl solution, the total volume becomes:
[
\text{Total Volume} = 100.0 \, \text{mL} + 100.0 \, \text{mL} = 200.0 \, \text{mL}
]
Step 3: Calculate the new concentration of HCl after dilution.
The moles of HCl remain constant before and after dilution. Initially, we have:
[
\text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.001 \, \text{M} \times 0.1 \, \text{L} = 0.0001 \, \text{moles}
]
After dilution, the concentration of HCl in the total volume of 200.0 mL is:
[
\text{New Concentration} = \frac{\text{Moles of HCl}}{\text{Total Volume}} = \frac{0.0001 \, \text{moles}}{0.2 \, \text{L}} = 0.0005 \, \text{M}
]
Step 4: Calculate the new pH.
Now we can find the new pH using the new concentration of hydrogen ions:
[
\text{pH} = -\log[H^+] = -\log(0.0005) \approx 3.30
]
Conclusion:
The resulting pH of the solution after adding 100.0 mL of water to the original 100.0 mL solution of hydrochloric acid with a pH of 3.00 is approximately 3.30. This demonstrates that dilution of an acid leads to an increase in pH, reflecting a decrease in hydrogen ion concentration.