Question

Your supervisor asks you to prepare 474.00 mL of ammonium buffer solution 0.1 M with pH= 9.00, from a concentrated ammonia solution (16%w/w, d=0.98g/mL) and solid ammonium chloride (97% w/w). (MW of NH3 = 17.031 g/mol and FW of NH4Cl = 53.491 g/mol and pKa NH4+/NH3 =9.24). How many grams of the solid ammonium chloride do you need? Consider two decimal places for the answer.

Answer 1

pka of NH4+= 9.24 pH of a buffer is determined by Henderson equation, pH- pk,+log [base] [acid] NH,] 9.00-924+10g NH, NH-0.24 log [NH4 [NH3] -10-0.57544 INH, ] INH,]-0.57544xNH] conc. of buffer - 0.1M That is [NH,]+[NH,-]= 0.1 0.57544x[NH,-]+[NH,-]=0.1 1.57544x[NH4+] = 0.1 -0.24 [NH,-]= [NH,]= 0.57544× [NH4+] = 0.57544 × 0.063474-0.036526 NH410.063474M [NH3-0036526M Volume of the buffer solution 474 mL 0.474 L Number of moles of [NH4 ] present in the solution -M*0.063474Mx0.474 L 0.030087 mol Number of moles of NH, present in the solution - M 0.036526Mx0.474 L 0.017313 mol Mass of NH,Cl moles*mol.Wt. 0.030087 molx 53.491 g/nol-1.609375 g %w/w of NH,CI = 97% 1.57544 =0.063474 x100 mass of solid NH CI mass of solid NH,CI mass of NH,C x100 mass of solid ŅH,C1 = 1.609375 mass of solid NH,Cl-1.66 g x 100-1.659149 g