An analytical chemist is titrating 216. 5 mL of a 0. 5900M solution of ammonia (NH3: ) with a 0. 4000M solution of HNO. The pKb of ammonia is 4. 74. Calculate the pH of the base solution after the chemist has added 375. 9 mL of the HNO3 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO3 solution added. Round your answer to 2 decimal place
To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the concentrations of the acid and its conjugate base.
The reaction between ammonia (NH3) and HNO3 can be represented as follows:
NH3 + HNO3 -> NH4+ + NO3-
The initial concentration of NH3 is 0.5900 M, and the final volume of the solution is the sum of the initial volume (216.5 mL) and the volume of HNO3 added (375.9 mL). Therefore, the final volume is 592.4 mL.
Next, we need to determine the concentrations of NH3 and NH4+ in the final solution after the addition of HNO3.
The moles of NH3 in the initial solution can be calculated as follows:
moles of NH3 = initial volume (L) * initial concentration (M)
= 0.2165 L * 0.5900 M
= 0.12771 moles
Since the molar ratio of NH3 to NH4+ is 1:1, the concentration of NH4+ in the final solution is also 0.12771 M.
Next, we need to determine the concentration of NO3- in the final solution. Since HNO3 is a strong acid, it completely dissociates to form H+ and NO3-. Therefore, the concentration of NO3- is equal to the concentration of HNO3, which is 0.4000 M.
Now, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution:
pH = pKa + log10([A-]/[HA])
Since NH3 acts as a base, NH4+ is its conjugate acid. The pKa of NH4+ can be calculated using the pKb of NH3:
pKa = 14 - pKb = 14 - 4.74 = 9.26
Substituting the values into the Henderson-Hasselbalch equation:
pH = 9.26 + log10(0.12771/0.12771)
= 9.26 + log10(1)
= 9.26
Therefore, the pH of the base solution after the addition of HNO3 is 9.26.
To calculate the pH of the base solution after adding the HNO3 solution, we need to determine the moles of NH3 and HNO3 reacted, and then calculate the concentration of NH4+ andOH- ions in the solution.
Step 1: Calculate the moles of NH3 in the 216.5 mL solution.
moles of NH3 = volume of solution (L) x concentration (mol/L)
moles of NH3 = 0.2165 L x 0.5900 mol/L
moles of NH3 = 0.12785 mol
Step 2: Calculate the moles of HNO3 added to the solution.
moles of HNO3 = volume of solution (L) x concentration (mol/L)
moles of HNO3 = 0.3759 L x 0.4000 mol/L
moles of HNO3 = 0.15036 mol
Step 3: Determine the limiting reactant.
For the reaction between NH3 and HNO3, the stoichiometry is 1:1. Therefore, the limiting reactant is NH3, as it has the smaller number of moles.
Step 4: Calculate the moles of NH4+ formed.
moles of NH4+ = moles of NH3 reacted = 0.12785 mol
Step 5: Calculate the moles of OH- formed.
moles of OH- = moles of NH4+ = 0.12785 mol
Step 6: Calculate the concentration of NH4+ and OH- ions.
concentration of NH4+ = moles of NH4+ / final volume (L)
concentration of NH4+ = 0.12785 mol / (0.2165 L + 0.3759 L)
concentration of NH4+ = 0.345 mol/L
concentration of OH- = moles of OH- / final volume (L)
concentration of OH- = 0.12785 mol / (0.2165 L + 0.3759 L)
concentration of OH- = 0.345 mol/L
Step 7: Calculate the pOH.
pOH = -log10(concentration of OH-)
pOH = -log10(0.345)
pOH = 0.461
Step 8: Calculate the pH.
pH = 14 - pOH
pH = 14 - 0.461
pH = 13.54
Therefore, the pH of the base solution after adding 375.9 mL of the HNO3 solution is 13.54.