I am completely lost with this question. If 50 ml of 0.5M HCl is added to 800 ml of pure water, what will be the approximate pH of the resulting solution?

First you must assume that the volumes are additive; i.e., the total volume will be 850

(HCl) = 0.5M x (50/850) = ?
Then pH = -log(HCl) = ?

Thank you so much!

To find the approximate pH of the resulting solution, we need to use the concept of dilution. Dilution refers to the process of reducing the concentration of a solute in a solution, usually by adding solvent. In this case, we are diluting the HCl by adding water.

To solve this question, we'll follow these steps:

Step 1: Determine the moles of HCl in the initial solution.
To find the moles of HCl, we multiply the molarity (M) by the volume in liters (V):
0.5 mol/L x 0.050 L = 0.025 mol

Step 2: Determine the moles of HCl in the final solution.
After adding 800 ml of water, the total volume becomes 50 ml + 800 ml = 850 ml. To convert this to liters, divide by 1000:
850 ml ÷ 1000 = 0.850 L

The number of moles remains the same, so the moles of HCl in the final solution is also 0.025 mol.

Step 3: Calculate the new concentration (M) of HCl in the final solution.
To find the new concentration, divide the moles of HCl by the volume in liters:
Concentration (M) = moles (mol) / volume (L)
0.025 mol / 0.850 L ≈ 0.029 M

Step 4: Convert the concentration (M) of HCl to pH.
The pH scale is a logarithmic scale that measures the acidity of a solution. To convert the concentration of HCl to pH, we need to calculate the negative logarithm (base 10) of the concentration:
pH = -log[H+]

For HCl, the concentration of H+ ions is the same as the concentration of HCl. Therefore, the pH is approximately:
pH = -log(0.029) ≈ 1.5

So, the approximate pH of the resulting solution after adding 50 ml of 0.5M HCl to 800 ml of pure water is approximately 1.5.