Question

a) Calculate the pH of a buffer that is 0.20 M NH3 and 0.20 M NH4Cl Correct?

b) Calculate the pH after addition of 10 mL 0.10 M HCl to 65 mL of the buffer?

c) Calculate the pH after addition of 5 mL of 1.0 M LiOH to 200 mL of the buffer?

Answer 1

INHUUT POH = pkt & log [NH₂] (NH4) = 14 [NH3) K6 for Ammonia (NH3) = 1.8x105 The initial pH of Buffer is- pH = 14-POH - 14-(pkb + log ENAM) pkb + log = 14-[-log (108x105) + log 0:20 ] = 14 - [4.744) 9 9.256. 10m 0.10 in He= Immols 65 m) X 0.2th NH₂ = 13 mmols NH3 es ml x 0.2 M NH 44 = 13 mmols NHYCE. The Reaction is – NH₃ H4 Initially 13 0 NHyu 13 a dd - ... ----- - -1...+1 12-0 ... 14 F Then, pH = pua & log 1149 0 1127

© 5 mit of t.om LiOH = m mols LiOH 2oom of 0.2 ds NH3 = 40 omal kHz 200 ml of 0.2 ds NHyu 40 mmolNHYCH AS LOH as a Barse con ond Ni also anno Base so they can be not be react to each other.