Question

How many grams of dry NH4Cl need to be added to 1.90 L of a 0.800 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.74? Kb for ammonia is 1.8×10^−5. Express your answer with the appropriate units.

Answer 1

weak base = NH3 salt - NH4CI Volume = 1.90 L NH3 = 0.800 M Kb = 1.8x10-5 pH = 8.74 molar mass of NH4CI = 53.49 g/mol NH4Cl = ? g calculating pkb: pKb = -log[Kb] pKb = -log[ 1.8x10-5 ] pkb = 4.74 calculating pOH: pOH = (14 - pH) pOH = (14 -8.74 ) pOH = 5.26

] calculating concentration: According to Henderson-Hasselbalch equation salt pOH = pKb + log ( weak base ] NH4Cl pOH = pKb + log NH3 5.26 = 4.74 + log NH4Cl 0.800 ] ] ] ] ] ] NH4Cl 0.52 log 0.800 NH4Cl = 2.62 M Calculating mass: mass Concentration = molar mass x Volume in L mass = ( Concentration molar mass ® Volume in L) mass =( 2.620 x 53.49 * 1.90 mass = 266.31 g added mass of NH4CI = 266.31 g