Answer:
The Kb of ammonia is 1.77 * 10^-5. The pH of a buffer prepared by combining 50.0 mL of 1.00 M ammonia and 50.0 mL of 1.00 M ammonium nitrate is 9.25
Step-by-step explanation:
Step 1: Data given
Kb of ammonia is 1.77 * 10^-5
Volume of ammonia is 50.0 mL = 50*10^-3L
Molarity of ammobia is 1.00 M
Volume of ammonium nitrate is 50.0 mL = 50*10^-3L
Molarity of ammonium nitrate is 1.00 M
Step 2: Calculate moles of ammoniumnitrate
Number of moles = molarity * volume
Number of moles NH4NO3 = 1 M * 0.05 L = 0.05 mol
Step 3: Calculate moles of ammonia
Number of moles = molarity * volume
Number of moles NH3 = 1 M * 0.05 L = 0.05 mol
Step 4: Calculate pH
pOH = - logKb + log [NH4NO3] /[NH3 ]
pOH= - log 1.77x10^-5 + log 1
= 4.752
Since pOh + pH = 14
pH = 14 - pOH = 14 - 4.752 = 9.248
pH ≈ 9.25
The Kb of ammonia is 1.77 * 10^-5. The pH of a buffer prepared by combining 50.0 mL of 1.00 M ammonia and 50.0 mL of 1.00 M ammonium nitrate is 9.25