Question

23. What is the pH of an aqueous solution made by combining 38.64 mL of a 0.4904 M ammonium chloride with 39.77 mL of a 0.3735 M solution of ammonia to which 4.670 mL of a 0.0772 M solution of HCl was added?

41. In a titration of 40.09 mL of 0.3795 M ammonia with 0.3795 M aqueous nitric acid, what is the pH of the solution when 40.09 mL of the acid have been added?

71. what is the molar solubility of barium fluoride?

Answer 1

23) pH of basic buffer =14 - (pkb+log(NH4Cl+HCl/NH3-HCl))

     pkb of NH3 = 4.75

no of mol of NH4Cl ADDED = 38.64*0.4904 = 18.95 mmol

no of mol of NH3 added = 39.77*0.3735 = 14.85 mmol

no of mol of Hcl added = 4.67*0.0772 = 0.36 mmol

pH = 14-(4.75+log((18.95+0.36)/(14.85-0.36))

    = 9.12

41)     pkb of NH3 = 4.75

no of mol of NH3 added = 40.09*0.3795 = 15.21 mmol

no of mol of HNO3 added = 40.09*0.3795 = 15.21 mmol

so that , now the system is at equivalence point.

concentration of salt = 15.21/(40.09+40.09)

pH = 7-1/2(pkb+logC)

     = 7-1/2(4.75+log0.19)

     = 4.98

71) Ksp of BAf2 = S*(2S)^2

           Ksp    = 4S^3

   (1.84*10^-7) = 4s^3

s = solubility of BaF2 = 3.58*10^-3 M