Calculate the pH of a solution prepared by mixing 250.mL of 0.174 molal aqueous HF (density=1.10g/mL) with 38.7g of an aqueous solution that is 1.50% NaOH by mass (density= 1.02g/mL).
To calculate the pH of the solution, we need to consider the dissociation of HF and the reaction between HF and NaOH.
First, let's find the moles of HF and NaOH:
1. Moles of HF:
We are given that the solution is 0.174 molal, which means the solution contains 0.174 moles of solute (HF) per kilogram of solvent (water). Since we have 250 mL of the solution and the density is 1.10 g/mL, the mass of the solution can be calculated as:
Mass of solution = Volume × Density
= 250 mL × 1.10 g/mL
= 275 g
Now let's calculate the moles of HF in the solution:
Moles of HF = Molality × Mass of solvent (water) in kg
= 0.174 mol/kg × 0.275 kg
= 0.04785 moles
2. Moles of NaOH:
We have 38.7 g of the 1.50% NaOH solution. Since the density of the solution is 1.02 g/mL, the mass of the solution can be calculated as:
Mass of solution = Volume × Density
= 38.7 g / 0.015
= 2580 g
Now let's calculate the moles of NaOH in the solution:
Moles of NaOH = Mass of NaOH / Molar mass of NaOH
= 38.7 g / 40.00 g/mol
= 0.9675 mol
Next, let's determine the reaction between HF and NaOH:
1. HF dissociation reaction:
HF, being a weak acid, partially dissociates in water:
HF (aq) ⇌ H+ (aq) + F- (aq)
2. Reaction between HF and NaOH:
HF (aq) + NaOH (aq) → NaF (aq) + H2O (l)
Since HF is a weak acid, we assume negligible H+ (aq) concentration before the reaction, and NaF is a strong electrolyte, meaning it will form F- ions.
Now, let's calculate the final volume of the solution after mixing:
Volume of HF solution = 250 mL
Volume of NaOH solution = Mass of NaOH solution / Density = 38.7 g / 1.02 g/mL = 37.94 mL
Final volume of the mixed solution = Volume of HF solution + Volume of NaOH solution
= 250 mL + 37.94 mL
= 287.94 mL
Now, let's calculate the molarity of the solution:
Molarity (M) = Moles / Volume (L)
= (moles of HF + moles of NaOH) / (volume of mixed solution in liters)
= (0.04785 moles + 0.9675 moles) / (287.94 mL / 1000)
= 3.67 M
Finally, let's solve for the pH of the solution using the formula:
pH = -log[H+]
Since HF partially dissociates in water, we need to consider the concentration of H+ from the dissociation of HF:
HF (aq) ⇌ H+ (aq) + F- (aq)
The concentration of H+ ions can be assumed equal to the concentration of HF (3.67 M).
Therefore, the pH of the solution is:
pH = -log[3.67]
= 0.435
Thus, the pH of the solution is approximately 0.435.