A sample of 7.80L of NH3 (ammonia) gas at 22 degrees and 735mmHg is bubbled into a 0.350L solution of 0.400M HCL(hydrochloric acid).
The Kb value for NH3 is 1.8x10-5
Assuming all the NH3 dissolves and that the volume of the solution remains at 0.350L , calculate the pH of the resulting solution.
To calculate the pH of the resulting solution, we need to find the concentration of NH4+ ions in the solution. This can be done by calculating the amount of NH3 that reacts with HCl and forms NH4+.
To start, let's calculate the moles of NH3 initially added to the solution. We can use the ideal gas law to find the number of moles of NH3:
PV = nRT
Where:
P = pressure of NH3 gas = 735 mmHg
V = volume of NH3 gas = 7.80 L
n = number of moles of NH3
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature in Kelvin = 22°C + 273.15 = 295.15 K
Rearranging the equation to solve for n, we have:
n = PV / RT
n = (735 mmHg * 7.80 L) / (0.0821 L·atm/(mol·K) * 295.15 K)
n = 21.57 mol
Now, let's calculate the moles of NH4+ ions produced when NH3 reacts with HCl. The balanced chemical equation for this reaction is:
NH3 + HCl -> NH4+ + Cl-
From the equation, we can see that one mole of NH3 reacts with one mole of HCl to form one mole of NH4+ ions. Therefore, the number of moles of NH4+ ions is also 21.57 mol.
Now, let's calculate the concentration of NH4+ ions in the solution. The volume of the solution remains at 0.350 L, and the number of moles of NH4+ ions is 21.57 mol. Therefore, the concentration (Molarity) can be calculated as follows:
Molarity = moles / volume
Concentration = 21.57 mol / 0.350 L
Concentration = 61.63 M
Now, let's determine the pOH of the solution. The pOH can be calculated using the equation:
pOH = -log[OH-]
The concentration of OH- ions can be calculated from the concentration of NH4+ ions using the Kb value for NH3. The Kb expression for the reaction between NH3 and water is:
Kb = [NH4+][OH-] / [NH3]
Since the concentration of NH4+ ions is the same as the concentration of OH- ions (due to 1:1 stoichiometry in the reaction), we can substitute [OH-] with [NH4+] in the Kb expression:
Kb = [NH4+][NH4+] / [NH3]
Plugging in the values:
1.8 x 10^-5 = (61.63 M)(61.63 M) / [NH3]
Solving for [NH3]:
[NH3] = (61.63 M)(61.63 M) / (1.8 x 10^-5)
[NH3] ≈ 2122463 M
Now, let's determine the pOH value:
pOH = -log[NH4+]
pOH = -log(61.63 M)
pOH ≈ 1.211
Finally, let's calculate the pH value of the solution:
pH = 14 - pOH
pH ≈ 14 - 1.211
pH ≈ 12.789
Therefore, the pH of the resulting solution is approximately 12.789.
To calculate the pH of the resulting solution, we need to determine the concentration of NH4+ (ammonium) ions in the solution after the reaction between NH3 and HCl.
1. Calculate the moles of NH3:
Volume of NH3 gas = 7.80 L
Using the ideal gas law: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L*atm/mol*K), and T is temperature in Kelvin.
Convert temperature from Celsius to Kelvin: 22 + 273.15 = 295.15 K
Determine pressure in atm: 735 mmHg / 760 mmHg/atm = 0.967 atm
Substitute these values into the ideal gas law: (0.967 atm)(7.80 L) = n(0.0821 L*atm/mol*K)(295.15 K)
Solve for n (moles of NH3).
2. Determine the number of moles of NH4+ ions produced:
Since the stoichiometry of the reaction between NH3 and HCl is 1:1, the number of moles of NH4+ ions produced will be equal to the number of moles of NH3.
3. Find the concentration of NH4+ ions:
Concentration (molarity) of NH4+ ions = moles of NH4+ ions / volume of solution (0.350 L)
4. Calculate the concentration of NH4+ ions in the solution:
Concentration (molarity) of NH4+ ions = moles of NH4+ ions / volume of solution (0.350 L)
5. Calculate the pOH of the solution:
Kb (equilibrium constant) for NH3 = [NH4+][OH-] / [NH3]
Since the concentration of NH3 is much larger than the concentration of NH4+ ions, we can assume that the concentration of NH3 is still approximately equal to the initial concentration of NH3.
Therefore, Kb = [NH4+][OH-] / [NH3]
Since we are assuming all NH3 dissolves, the concentration of NH3 is equal to the initial concentration.
Rearrange the equation to solve for [OH-]: [OH-] = (Kb * [NH3]) / [NH4+]
6. Find the pOH of the solution:
pOH = -log10([OH-])
7. Calculate the pH of the solution:
pH = 14 - pOH
Follow these steps to calculate the pH of the resulting solution.