Calculate the pH of a solution prepared by mixing 250. mL of 0.174 m aqueous HF (density = 1.10 g/mL) with 25.7 g of an aqueous solution that is 1.50% by mass (density = 1.02 g/mL).
I dont understand how to set it up
1.50% by mass what?
To calculate the pH of the solution, we need to determine the concentration of the hydrogen ion (H+) in the solution. We can use the equation:
pH = -log[H+]
To determine the concentration of H+ in the solution, we need to calculate the moles of the solute (HF) and the volume of the solution. Here's how to set it up:
Step 1: Calculate the moles of HF:
Moles of HF = concentration (molarity) × volume (in liters)
= 0.174 mol/L × (250 mL/1000 mL/L)
= 0.174 mol/L × 0.250 L
= 0.0435 mol
Step 2: Calculate the moles of water (H2O) in the additional solution:
Mass of water = density × volume
= 1.02 g/mL × (25.7 g / 1.02 g/mL)
= 25.7 g
Moles of water = mass / molar mass
= 25.7 g / 18.02 g/mol (molar mass of water)
≈ 1.43 mol
Step 3: Calculate the total volume of the solution:
Total volume = volume of HF solution + volume of additional solution
= 250 mL + (25.7 g / 1.02 g/mL)
= 250 mL + 25.2 mL
= 275 mL
Step 4: Calculate the concentration (molarity) of the HF solution after it is diluted with the additional solution:
Concentration of HF = moles of HF / total volume (in liters)
= 0.0435 mol / (275 mL / 1000 mL/L)
= 0.0435 mol / 0.275 L
= 0.158 M
Step 5: Calculate the concentration of H+ ions from HF:
Since HF is a weak acid, we consider it partially ionized in water:
HF ⇌ H+ + F-
The concentration of H+ ions is the same as the concentration of the acid, which is 0.158 M.
Step 6: Calculate the pH using the equation:
pH = -log[H+]
= -log(0.158)
≈ 0.8
Therefore, the pH of the solution is approximately 0.8.