Pure water is added to 25.0 mL of a 1.00 M HCL solution. The new solution has a volume of 2.00 L. What is the pH of the new solution.
1.00 M x (25.0 mL/2,000 mL) = (HCl)
pH = -log(HCl) = ?
To find the pH of the new solution after adding pure water, we need to calculate the concentration of the HCl solution after the dilution.
We can use the formula for dilution:
C1V1 = C2V2
Where:
C1 = initial concentration of the HCl solution
V1 = initial volume of the HCl solution
C2 = final concentration of the HCl solution
V2 = final volume of the solution (including the addition of water)
Given:
C1 = 1.00 M
V1 = 25.0 mL = 0.025 L
V2 = 2.00 L
Let's substitute the values into the formula:
(1.00 M)(0.025 L) = C2(2.00 L)
Solving for C2, the final concentration of the HCl solution, we get:
C2 = (1.00 M)(0.025 L) / (2.00 L)
C2 = 0.0125 M
Now that we know the final concentration of the HCl solution, we can calculate the pH using the equation for the pH of a strong acid solution:
pH = -log10(C)
Where C is the concentration of the HCl solution. Plugging in the value of C2:
pH = -log10(0.0125 M)
pH ≈ 1.9
Therefore, the pH of the new solution is approximately 1.9.