What is the pH of a solution prepared by dissolving 6.0 g of HCl in water to make 415 mL of HCl solution?
moles HCl = grams/molar mass.
Then M = moles/L soln.
To find the pH of a solution prepared by dissolving HCl in water, we need to know the concentration of HCl in the solution.
First, let's calculate the number of moles of HCl in the solution:
Molar mass of HCl = 1.00784 g/mol (H) + 35.453 g/mol (Cl) = 36.46084 g/mol
Number of moles of HCl = mass of HCl / molar mass of HCl
Number of moles of HCl = 6.0 g / 36.46084 g/mol = 0.164 moles
Next, we need to calculate the concentration of HCl in the solution (in moles per liter or Molarity):
Concentration (Molarity) = moles of solute / volume of solution in liters
Volume of solution = 415 mL = 415 mL / 1000 mL/L = 0.415 L
Concentration of HCl = 0.164 moles / 0.415 L ≈ 0.395 M
Now, we can use the concentration of HCl to find the pH of the solution. The pH scale is a logarithmic scale ranging from 0 to 14, where a pH of 7 is considered neutral, values below 7 are acidic, and values above 7 are basic.
The pH of a strong acid like HCl can be calculated using the formula:
pH = -log10[H+]
Since HCl is a strong acid, it fully dissociates in water to release hydrogen ions (H+). The concentration of H+ ions will be the same as the concentration of HCl in the solution.
pH = -log10(0.395)
pH ≈ -(-0.4038)
pH ≈ 0.4038
Therefore, the pH of the solution prepared by dissolving 6.0 g of HCl in water is approximately 0.4038.