Calculate OH in the following solutions A 60x103M BaOH2 and

Calculate [OH-] in the following solutions.

A. 6.0x10^-3M Ba(OH)2 and 1.00x10^-2M BaCl2

Express your answer using two significant figures.

B. 0.320M (NH4)2SO4 and 0.480M NH3

Express your answer using two significant figures.

C. 0.196M NaOH and 0.264M NH4Cl

Express your answer using two significant figures.

Solution

A. Kb = 10^-2.02 = 0.00955 Ba(OH)2 ---> Ba2+ + 2OH- 0.006-x............x.......... 2x kb = 0.00955 = x*(2x)^2/(0.006-x) x = 5.88*10^-3 [OH-] = 2x = 1.176*10^-2M ......................... B Kb of NH3 is 1.80x10-5 NH3 + H2O ---> NH4+ + OH- 0.48-x.................x..........x 1.8*10^-5 = x^2/(0.48-x) x = 2.93*10^-3M =[OH-] ............................... C NaOH ----> Na+ + OH- kb of NaOH is very large so all of NaOH is dissociated to Na+ and OH- so [OH-] = 0.196M