Consider the titration of 30.0 mL of 0.050M NH3 with 0.025M HCl . Calculate the

Question

Consider the titration of 30.0 mL of 0.050M NH3 with 0.025M HCl . Calculate the pH after the following volumes of titrant have been added. a. 60.0 mL b. 71.8 mL c. 73.3 mL

Explanation / Answer

NH3 + H2O -----> NH4+ OH-- , 0.05-x x x , Kb =(x^2)/(0.05-x) = 1.77 x10^ -5 (taken from wiki) , x= 0.000934 =[NH3] , moles of NH3 = (0.000934 x30/1000) = 0.000028, a) 60 ml HCl = 60 x0.025/1000 = 0.0015, net H+ moles = 0.0015-0.000028 = 0.001472, [H+] = (0.001472 x1000)/(30+60) = 0.01635, pH = -log(0.01635) = 1.786 b) 71.8 ml HCl , H+ = 71.8 x0.025/1000 = 0.001795, net H+ = 0.001795-0.000028 = 0.001767, [H+] = 0.001767 x 1000/(30+71.8) =0.017357 , pH = 1.76 c) 73.3 ml HCl , H+ = (73.3 x0.025/1000) = 0.0018325, net H+ = 0.0018325-0.000028 = 0.0018045, [H+]= 0.0018045 x1000/(30+73.3) = 0.01774 , pH = 1.75

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