Chemical Concentration Na2CO3 -> 1.00E-04 NaOCl -> 1.00E-03 NH4Cl -> 2.00E-03 So

Question

Explanation / Answer

55 = Na2CO3 , pkb = 4.67 , let Na2CO3 be BOH, ( since Na2CO3 + H2O --> H2CO3 + OH-) , BOH ---> B+ + OH- , [BOH] = 0.0001M , [B+]=[OH-] = 0.0001-x , Kb = 10^ ( -4.67) = [B+][OH-]/[BOH] , 2.14 x10^ -5 = x^2/( 0.0001-x) , x = 3.67 x10^ -5 , = [OH-] , pOH = -log( 3.67 x10^ -5) = 4.435 , pH = 14-4.435 = 9.565, [OH-]=[H2CO3] = 3.67 x10^ -5 , [Na2CO3] = 0.0001-0.0000367 = 6.33 x10^ -5 , for NaOCl , Kb = 3.14 x10^ -7 = x^2/(0.001-x) , x = [OH-] = 1.756 x 10^-5 , pOH = -log( 1.756 x10^ -5) = 4.755 , pH = 14-4.755 = 9.245, [NaOCl] = 0.001-0.00001756 =9.824 E-4 , for NH4Cl pkb = 4.755 , Kb = 10^ -4.755 = 1.757 x10^ -5 = x^2/(0.002-x) , x = 0.000178 = [OH-] , pOH = 3.745, pH = 10.255 , [NH4Cl] = 0.002-0.000178 = 0.0018

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.