c. Calculate the Kb for NH_3 given that a 0.05 M solution of NH4Cl has an observ

Question

c. Calculate the Kb for NH_3 given that a 0.05 M solution of NH4Cl has an observed pH of 5.20. Consider the given information collected during the titration of a sample of an oxalate containing barium oxalate as the soluble component, with 0.0010 M KMnO_4 solution. The burette contains the KMnO_4 solution. A 50 mL sample of the oxalate solution is used and 2 mL of HNO_3 is added to make the medium acidic. The balanced equation is: 2MnO_4^-1 + 5C_2 O_4^2- + 16H^+ rightarrow 2Mn2^+ + 10CO_2 + 8H_2 O Fill in the missing data using the information provided. Show the calculations clearly. To calculate [C_2 O_4^2-], use the total volume (50 + 2 + 7.46)mL

Explanation / Answer

NH4Cl --> NH4+ + Cl-

NH4+ + H2O <-> NH3 + H3O+

Ka = [NH3][H3O+]/[NH4+]

in equilibrium

[NH3] = [H3O+] = x

[NH4+] = 0.05-x

x can be calculated via

[H+] = 10^-pH = 10^-5.20 = 6.3095*10^-6 M = x

[NH3] = [H3O+] = x = 6.3095*10^-6

[NH4+] = 0.05-x = 0.05- 6.3095*10^-6

substitute and solve for Ka

Ka = (6.3095*10^-6)(6.3095*10^-6)/(0.05-6.3095*10^-6) = 7.9629*10^-10

Kb = Kw/a = (10^-14)/(7.9629*10^-10) = 1.255*10^-5

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