A technician requires a 4L of an ammonia solution to have a pH of 11.55. What volume in liters, of ammonia gas at 25 degrees Celsius and 1 atm is necessary to achieve this result?
NH3 + HOH ==> NH4^+ + OH^-
Kb = (NH4^+)(OH^-)/(NH3)
You want pH = 11.55 so pOH is 14-11.55 = ??. Convert pOH to OH^-, substitute for (OH^-) and for (NH4^+) in the Kb expression, and solve for NH3. That will give you (NH3) in moles/L. You will want 4 times that to make a 4 L solution, then plug those moles into PV = nRT and solve for volume at the conditions listed.
To determine the volume of ammonia gas required to achieve a specific pH in a 4L ammonia solution, we need to use the Henderson-Hasselbalch equation. The equation relates the pH of a solution to the concentration of the acid and its conjugate base.
The Henderson-Hasselbalch equation is:
pH = pKa + log([A-]/[HA])
In this case, ammonia (NH3) acts as a base and its conjugate acid is the ammonium ion (NH4+). The pKa value for the ammonium ion is known to be 9.25.
Step 1: Calculate the concentration of NH4+ (HA)
Since the volume of an ammonia solution is not given, we can assume it to be 4L. Therefore, the concentration of NH4+ (HA) can be calculated using the formula:
[HA] = 10^(-pH) * [A-]
[HA] = 10^(-11.55) * [A-]
Step 2: Convert the concentration of NH4+ (HA) to moles per liter (M)
The molar mass of NH4+ is 18.04 g/mol. Assuming 4L of ammonia solution, we can calculate the moles of NH4+ (HA) using the formula:
M = (moles of solute) / (volume of solution)
M = ([HA] * molar mass of NH4+) / volume of solution
Step 3: Convert moles of NH4+ (HA) to moles of NH3
Since NH3 and NH4+ have the same moles but different volumes, we need to convert moles of NH4+ (HA) to moles of NH3. This can be done using the stoichiometry of the reaction:
NH4+ (aq) + OH- (aq) -> NH3 (g) + H2O (l)
Step 4: Convert moles of NH3 to volume in liters
Finally, we can convert the moles of NH3 to volume in liters using the ideal gas law:
PV = nRT
Here, we assume the pressure is 1 atm, the temperature (T) is given as 25 degrees Celsius (298 K), and R is the ideal gas constant (0.0821 L*atm/(mol*K)).
Now, let's calculate the volume of ammonia gas required to achieve a pH of 11.55 in a 4L ammonia solution.
Note: The calculations will involve several steps. If you'd like, I can provide the numerical calculations for you.
To solve this problem, we need to use the concept of the ideal gas law and the equation for the dissociation of ammonia in water.
The first step is to determine the amount of ammonia in moles needed to achieve the desired pH of the ammonia solution. We can calculate this using the equation:
pOH = 14 - pH
Since pH + pOH = 14, we can determine the pOH value:
pOH = 14 - 11.55 = 2.45
To convert pOH to concentration of hydroxide ions ([OH-]), we use the equation:
pOH = -log[OH-]
Taking the antilog of both sides, we get:
[OH-] = 10^(-pOH) = 10^(-2.45)
[OH-] = 0.003548 moles/L
Since the dissociation of ammonia in water is written as:
NH3 + H2O ⇌ NH4+ + OH-
And the stoichiometry of the reaction is 1:1, the concentration of ammonia would also be 0.003548 moles/L.
The next step is to use the ideal gas law, which states:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
We can rearrange the ideal gas law equation to solve for volume:
V = nRT / P
Now, let's substitute the given values:
n = 0.003548 moles
R = 0.0821 L.atm/mol.K
T = 25°C + 273.15 (converting Celsius to Kelvin) = 298.15 K
P = 1 atm
Plugging these values into the equation, we get:
V = (0.003548 moles)(0.0821 L.atm/mol.K)(298.15 K) / 1 atm
V = 0.0896 L
Therefore, the volume of ammonia gas at 25°C and 1 atm necessary to achieve a pH of 11.55 for a 4L ammonia solution is approximately 0.0896 liters.