10 mL of 0.1 HCl was added to 9mL of water. 10 mL of that solution was added to another 90 mL of water. The student diluted the sample as above 1 time(s). What is the pH of the solution?
first dilution: you diluted it to 10/19 of .1M, so it is 10/190 M.
Then you diluted it 10/100 of its concentration, new concentration 10/1900 M
the statement <The student diluted the sample as above 1 time(s).>> makes no sense to me, it it means my first paragraph, then the concentration is 10/1900 M
pH=-log(10/1900)=2.28
check all that.
To determine the pH of the solution, we need to calculate the final concentration of the hydrochloric acid (HCl) after the dilution.
Step 1: Determine the initial concentration of the HCl solution.
We have 10 mL of 0.1 M HCl. This means that there are 0.1 moles of HCl in 1 liter (1000 mL) of the solution.
Step 2: Calculate the amount of moles in the 10 mL of 0.1 M HCl.
1 liter of 0.1 M HCl contains 0.1 moles, so 10 mL of 0.1 M HCl contains (0.1/1000) moles of HCl.
moles HCl = (0.1/1000) moles HCl = 0.0001 moles HCl
Step 3: Calculate the final volume of the solution after dilution.
10 mL of the initial solution was mixed with 90 mL of water, resulting in a total volume of 100 mL.
Step 4: Calculate the final concentration of HCl after dilution.
The final concentration of HCl can be calculated as follows:
(final moles HCl) / (final volume of solution in liters) = (0.0001 moles) / (0.1 liters) = 0.001 moles/L
Step 5: Calculate the pH of the solution.
The pH of a solution can be calculated using the following formula:
pH = -log[H+]
Since HCl is a strong acid, it completely ionizes in water, meaning that the concentration of H+ ions is equal to the concentration of HCl.
pH = -log(0.001) = 3
Therefore, the pH of the solution after the dilution is 3.
To determine the pH of the solution, we need to consider the dilution process. Let's break down the steps:
1. First, 10 mL of 0.1 M HCl was added to 9 mL of water. This creates a diluted solution. We can calculate the concentration of the resulting solution using the formula:
C₁V₁ = C₂V₂
where C₁ is the initial concentration, V₁ is the initial volume, C₂ is the final concentration, and V₂ is the final volume.
Substituting the given values, we have:
(0.1 M)(10 mL) = C₂(10 mL + 9 mL)
Solving for C₂:
C₂ = (0.1 M)(10 mL) / (19 mL)
≈ 0.0526 M
2. Next, 10 mL of the diluted solution (C₂) is added to 90 mL of water. This is another step of dilution.
Repeating the dilution calculation, we have:
(C₂)(10 mL) = C₃(10 mL + 90 mL)
Solving for C₃:
C₃ = (C₂)(10 mL) / (100 mL)
= (0.0526 M)(10 mL) / (100 mL)
≈ 0.00526 M
So, the final concentration after the second dilution is approximately 0.00526 M.
3. Finally, to determine the pH, we can use the fact that HCl is a strong acid, meaning it completely ionizes in water. Therefore, the concentration of H⁺ ions in the solution is equal to the concentration of the HCl solution.
pH = -log[H⁺]
= -log(0.00526 M)
≈ 2.28
Therefore, the pH of the solution after the dilution is approximately 2.28.