What is the pH of a solution that results from the addition of 5mL of 0.1 M HCl to 50mL of pure water?

To determine the pH of the resulting solution, we need to calculate the concentration of H+ ions after the HCl is added. The concentration of HCl can be calculated using the formula:

Concentration (C) = moles (n) / volume (V)

Given:
Volume of HCl (V1) = 5 mL = 0.005 L
Concentration of HCl (C1) = 0.1 M

Using the formula, we can calculate the moles of HCl:

Moles of HCl = C1 * V1

Moles of HCl = 0.1 M * 0.005 L

Moles of HCl = 0.0005 moles

Now, we need to calculate the total volume of the resulting solution after the HCl is added. This can be calculated as:

Total Volume (Vt) = Volume of HCl (V1) + Volume of Water (V2)

Total Volume (Vt) = 0.005 L + 0.050 L

Total Volume (Vt) = 0.055 L

Next, we need to calculate the concentration of H+ ions in the resulting solution. Since the HCl is a strong acid, it dissociates completely in water, meaning that the moles of H+ ions equal the moles of HCl:

Moles of H+ ions = Moles of HCl = 0.0005 moles

Finally, we can calculate the concentration of H+ ions in the resulting solution:

Concentration of H+ ions (CH+) = Moles of H+ ions / Total Volume (Vt)

Concentration of H+ ions (CH+) = 0.0005 moles / 0.055 L

Concentration of H+ ions (CH+) ≈ 0.0091 M

Since we know that pH is defined as the negative logarithm (base 10) of the H+ concentration, we can calculate the pH of the resulting solution:

pH = -log(CH+)

pH = -log(0.0091)

pH ≈ 2.04

Therefore, the pH of the resulting solution is approximately 2.04.

mL x M = mL x M

Solve for the M of the diluted HCl, then plug into the pH equation.
Then pH = -log(H^+) = ??