A particular household ammonia solution (d=0.97 g/mL) is 6.8% NH3 by mass. How many mililiters of this solution should be diluted with water to produce 625mL of a solution with pH = 11.65.
Is that 6.8% by mass (w/w) or by volume. I will assume it is w/w which means 6.8g/100 g soln. Convert 6.8g NH3 to moles (moles = grams/molar mass) and convert 100 g soln to volume (mass = volume x density). Then calculate M NH3 = moles/L soln. That will give you the molarity of the solution you have.
Next calculate the molarity of the soln of the diluted material.
You want pH = 11.65. Calculate pOH (pH + pOH = pKw = 14) then use
NH3 + H2O ==> NH4^+ + OH^-
Set up an ICE chart for NH3. Substitute OH^- needed and solve for NH3 initially. That will allow you to use mL 6.8%soln x M6.8%soln = 625 x M NH3 needed. Solve for mL6.8%soln.
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To determine the volume of the ammonia solution that should be diluted with water, we first need to calculate the mass of NH3 in the final solution.
Step 1: Calculate the mass of NH3 in 625 mL of solution with pH = 11.65.
We need to use the pH value to calculate the concentration of NH3 in the final solution.
Step 2: Calculate the concentration of NH3 in the final solution.
The concentration of NH3 can be determined using the pKb value. The pKb value is the negative logarithm of the base dissociation constant (Kb) of a substance. In this case, NH3 acts as a base and can be written as:
NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Given: pH = 11.65
Since pH + pOH = 14,
pOH = 14 - 11.65 = 2.35
pOH = -log[OH-]
2.35 = -log[OH-]
Taking the anti-log of both sides:
[OH-] = 10^-2.35
[OH-] ≈ 0.004466
We know that [OH-] = [NH4+], so the concentration of NH4+ is also 0.004466 M.
The Kb for NH4+ can be calculated using the following equation:
Kb = Kw / Ka
Where Kw is the ion product of water (1.0 x 10^-14 at 298K), and Ka is the acid dissociation constant.
Given: Kw = 1.0 x 10^-14
Kb = (1.0 x 10^-14) / (Ka)
Kb ≈ 1.0 x 10^-14
Now we can calculate the concentration of NH3 using the Kb value:
Kb = [NH3][OH-]
1.0 x 10^-14 = [NH3] * 0.004466
Solving for [NH3]:
[NH3] = (1.0 x 10^-14) / 0.004466
[NH3] ≈ 2.24 x 10^-12
Step 3: Calculate the mass of NH3 in 625 mL of solution.
Mass of NH3 = Volume of solution * Concentration of NH3
Mass of NH3 = 0.625 L * 2.24 x 10^-12 mol/L * 17.03 g/mol (molar mass of NH3)
Mass of NH3 ≈ 2.25 x 10^-9 g
Step 4: Calculate the mass of the ammonia solution needed.
We know that the ammonia solution is 6.8% NH3 by mass. This means that for every 100 g of solution, 6.8 g is NH3.
Let x be the mass of the ammonia solution needed. We can set up the following equation:
2.25 x 10^-9 g = (6.8 g / 100 g) * x
x = (2.25 x 10^-9 g) / (6.8 g / 100 g)
x ≈ 3.31 x 10^-8 g
Step 5: Calculate the volume of the ammonia solution needed.
Density (d) = Mass / Volume
0.97 g/mL = (3.31 x 10^-8 g) / Volume
Volume = (3.31 x 10^-8 g) / (0.97 g/mL)
Volume ≈ 3.41 x 10^-8 mL
Therefore, approximately 3.41 x 10^-8 mL of the ammonia solution should be diluted with water to produce 625 mL of a solution with pH = 11.65.
To find the number of milliliters of the household ammonia solution that should be diluted, we need to calculate the concentration of NH3 in the final solution.
First, we can calculate the mass of NH3 present in 625 mL of solution with a pH of 11.65.
pH is a measure of the concentration of hydrogen ions (H+). In a basic solution, the pH value is greater than 7, indicating a low concentration of H+ ions.
Ammonia (NH3) acts as a base and reacts with water to produce ammonium ions (NH4+) and hydroxide ions (OH-). The increase in the concentration of hydroxide ions leads to a higher pH value.
The pOH is the negative logarithm of the hydroxide ion concentration. Using the pOH value, we can calculate the concentration of OH- ions.
pOH = 14 - pH
pOH = 14 - 11.65
pOH ≈ 2.35
The concentration of OH- ions can be calculated using the pOH value:
OH- concentration = 10^(-pOH)
OH- concentration = 10^(-2.35)
Next, we need to consider the balanced equation for the reaction of NH3 with water:
NH3 + H2O ↔ NH4+ + OH-
From the balanced equation, we can determine that the concentration of OH- ions is equal to the concentration of NH3 since each molecule of NH3 that reacts with water produces one molecule of OH-.
Thus, the concentration of NH3 is also approximately 10^(-2.35) M.
Now, we need to find the amount of NH3 (mass) in the 625 mL solution.
Concentration of NH3 (in g/L) = Concentration of NH3 (in M) × molar mass of NH3 (in g/mol) = 10^(-2.35) × (17 g/mol)
Next, we can calculate the mass of NH3 present:
Mass of NH3 (in g) = Concentration of NH3 (in g/L) × Volume of solution (in L)
Since concentration is usually expressed in grams per milliliter (g/mL), we convert the volume from milliliters to liters:
Volume of solution (in L) = 625 mL ÷ 1000 mL/L
Finally, we can calculate the mass of NH3:
Mass of NH3 (in g) = (10^(-2.35) × (17 g/mol)) × (625 mL ÷ 1000 mL/L)
Now, we need to find the amount of the household ammonia solution that contains this mass of NH3.
The solution is 6.8% NH3 by mass, so the mass of the household ammonia solution can be determined using the following equation:
Mass of household ammonia solution (in g) = Mass of NH3 (in g) ÷ (NH3 mass % ÷ 100)
Mass of household ammonia solution (in g) = Mass of NH3 (in g) ÷ (6.8% ÷ 100)
Next, we can calculate the volume of the household ammonia solution in milliliters using the density of the solution:
Volume of household ammonia solution (in mL) = Mass of household ammonia solution (in g) ÷ Density of solution (in g/mL)
Finally, to find the number of milliliters of this solution that should be diluted with water to produce 625 mL of the final solution, subtract the volume of the household ammonia solution from 625 mL:
Volume of water (in mL) = 625 mL - Volume of household ammonia solution (in mL)
This calculation will give you the number of milliliters of the household ammonia solution that should be diluted with water to produce 625 mL of a solution with a pH of 11.65.