Question

38. A buffer solution is 0.40 M NH3 and 0.60 M NHACI. Kb for NH3 is 1.8 x 10-5 a) Calculate the pH for the buffer system. b) Calculate the pH of the solution after adding 0.00600 moles of NaOH to 400.0 mL of the buffer. c) Calculate the pH of the solution after adding 0.050 moles of HCl to 600.0 mL of the buffer.

Answer 1

39) 4.7442 Kt or NH3 = To Kio-5 pkt = -log (1.68x10-5)= (NH33 = 0.40M сунусор обо м PH= 14- (PRG +log Hure 0.40 = 14-(4.7447 +log 0:50) = 14-(4.7447 4 0.1761) = 14-ugro PH = g.og no. of moles of NaoH= 0.005 vol of butter = 400 m2 0.42 no. of mole of NHG= 0.40 M X 0.402 = 0.15 hole to a m o le sg #Hung = 0 So Yo | 0 = 0.Ru mole ho. od melc od MHz astor NGOH = 0.16 +000G = 0.166 mole no. of mole a Nhyce after Haon = 0.2u - 01006 59234 - 1u-( 4.744 7 + log 0.734 0.166 14- ( 4.7447 +0.1491) 14 - 4.8939 PH = 9, 1052 g || DH = 9 )

addition of 0.050 moles of tre volt tuz fer = 0.602 no. of mole of NH₃= 0.40xo.ro = 0.24 mole noi o mole og NHUO = 0.50 0.50 = 0.36 mole PH= 14- (PKG +10g Nhuce) MHS ) no. of mole of NH3 after the - a2u oo to = 0.110 = 0.24 -0.05 = 0.19 mole no. of mole of HHUTE = 0.35 to.05 = D.ul mole PH= 14- 4.7447 & log a olg 0.41 = 14-14.3947 +0:334) 14-510757 PH=Giga