A buffered solution is made by adding 50.0 g NH4Cl to 1.00 L of a 0.84 M solution of NH3. Calculate the pH of the final solution
jemal
To calculate the pH of the final solution, we need to determine the concentration of NH4+ and NH3 in the solution and apply the Henderson-Hasselbalch equation.
First, let's find the moles of NH3. We can use the formula:
moles = concentration × volume
moles of NH3 = 0.84 M × 1.00 L = 0.84 moles
Next, let's find the moles of NH4Cl. We can use the formula:
moles = mass / molar mass
molar mass of NH4Cl = 14.01 g/mol (molar mass of NH3) + 1.01 g/mol (molar mass of H) + 35.45 g/mol (molar mass of Cl) = 53.47 g/mol
moles of NH4Cl = 50.0 g / 53.47 g/mol = 0.936 moles
Now, let's compare the moles of NH3 and NH4+. If the moles of NH4+ are greater than the moles of NH3, the excess NH4+ will react with water to form NH3 and H3O+. However, if the moles of NH4+ are lesser than the moles of NH3, we need to calculate the concentration of NH4+ and NH3.
In this case, the moles of NH4+ are equal to the moles of NH3:
moles of NH4+ = moles of NH3 = 0.84 moles
Since the total volume is 1.00 L, the concentration of NH4+ and NH3 is:
concentration = moles / volume
concentration of NH4+ = concentration of NH3 = 0.84 moles / 1.00 L = 0.84 M
Now, we can apply the Henderson-Hasselbalch equation:
pH = pKa + log [A-] / [HA]
In this case, the NH3 acts as the base (A-) and NH4+ acts as the acid (HA). The pKa of the NH4+/NH3 system is 9.25 (at 25 °C).
pH = 9.25 + log [0.84 M] / [0.84 M]
pH = 9.25 + log 1
pH = 9.25
Therefore, the pH of the final solution is 9.25.
To calculate the pH of the final solution, we need to determine the concentration of the hydronium ions (H3O+) in the solution.
First, we need to find the moles of NH4Cl in the solution. To do this, we divide the given mass of NH4Cl by its molar mass:
Molar mass of NH4Cl = 14.01 g/mol (N) + 1.01 g/mol (H) * 4 + 35.45 g/mol (Cl) = 53.49 g/mol
Moles of NH4Cl = 50.0 g / 53.49 g/mol = 0.935 mol
Next, we need to find the moles of NH3 present in the solution. To do this, we multiply the volume of the NH3 solution by its molarity:
Moles of NH3 = 1.00 L x 0.84 mol/L = 0.84 mol
Now, let's consider the reaction between NH4Cl and NH3:
NH4Cl(aq) + NH3(aq) ⇌ NH4+(aq) + Cl-(aq)
Since NH4Cl is a strong acid and NH3 is a weak base, NH4Cl will completely ionize in water, producing NH4+ and Cl- ions. Therefore, the concentration of NH4+ ions in the solution is equal to the moles of NH4Cl. The concentration of NH3 is the same as its initial molarity.
Now we can set up the Henderson-Hasselbalch equation to calculate the pH of the buffered solution:
pH = pKa + log([A-]/[HA])
In this case, NH4+ is the acid (HA) and NH3 is the base (A-). The pKa of NH4Cl is known to be 9.25.
We substitute the given values into the equation:
pH = 9.25 + log([NH3]/[NH4+])
pH = 9.25 + log(0.84 mol/0.935 mol)
Finally, we can calculate the pH:
pH = 9.25 + log(0.898)
Using a calculator, we find:
pH ≈ 9.25 + (-0.046)
pH ≈ 9.204
Therefore, the pH of the final solution is approximately 9.204.