A particular household ammonia solution (d= 0.97g/mL) is 6.8% NH3 by mass.
How many milliliters of this solution should be diluted with water to produce 650 mL of a solution with pH = 11.55?
Use the same procedure of the KOH problem I showed you earlier.
To solve this problem, we need to follow a step-by-step approach:
Step 1: Calculate the mass of ammonia (NH3) in the initial solution.
First, we need to know the mass of the original solution. Since we are given the density (d=0.97g/mL) and we want to produce 650 mL of the final solution, we can calculate the mass of the original solution using the formula:
mass = volume × density
mass = 650 mL × 0.97 g/mL
mass = 630.5 g
Now, we need to determine the mass of ammonia in the original solution. Since the solution is 6.8% NH3 by mass, we can calculate the mass of ammonia (NH3) as follows:
mass of NH3 = 6.8% × mass of the original solution
mass of NH3 = (6.8/100) × 630.5 g
mass of NH3 = 42.874 g
So, the mass of ammonia in the original solution is 42.874 g.
Step 2: Determine the concentration of NH3 in the final solution.
Since we want to produce a solution with pH = 11.55, we know that this pH value corresponds to a basic solution. Ammonia is a weak base, so we can use the pKa value of ammonia (NH3) to calculate its concentration.
The pKa value of ammonia is 9.25. We can use the Henderson-Hasselbalch equation to calculate the concentration of NH3 in the final solution:
pH = pKa + log10 ([A-]/[HA])
Since ammonia acts as a base, the concentration of NH3 (A-) is what we want to determine, and the concentration of NH4+ (HA) is water (H2O), which is taken as approximately 55.5 M.
11.55 = 9.25 + log10 ([NH3]/[H2O])
Subtracting 9.25 from both sides:
2.30 = log10 ([NH3]/[H2O])
To eliminate the logarithmic form, we can rewrite the equation as:
[NH3]/[H2O] = 10^(2.30)
[NH3]/[H2O] = 1995.262
Since water is in excess when diluting the initial solution, we can assume the concentration of water remains constant.
Step 3: Calculate the volume of the original solution that needs to be diluted.
Now that we know the concentration of NH3 in the final solution and the mass of NH3 in the original solution, we can determine the volume of the original solution that needs to be diluted.
[NH3] = mass of NH3 / mass of solution
mass of solution = mass of NH3 / [NH3]
mass of solution = 42.874 g / 1995.262
mass of solution = 0.0215 g
Now, use the density (d=0.97g/mL) to find the volume of the original solution:
volume of solution = mass of solution / density
volume of solution = 0.0215 g / 0.97 g/mL
volume of solution ≈ 0.0222 mL
So, approximately 0.0222 mL of the original solution needs to be diluted with water to produce 650 mL of a solution with pH = 11.55.