Question

AGID a)Calculate the pH of a buffer solution which is 0.30 M CH3CH2COH and 0.20 M NACH, CH, CO2 b) Calculate the pH after 0.0
I'm stuck in part (b) and part (c). If you could please explain both I would appreciate it. thank you in advance :)
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Answer #1

(b). Addition of 0.05M NaOH :-

ICF table is :

....................................CH3COOH (aq)..........+.............NaOH (aq) --------------> CH3COONa (aq)...........+.............H2O (l)

Initial ..............................0.30 M......................................0.05 M...............................0.20 M...............................................

Change...........................-0.05 M.....................................-0.05 M..............................+0.05 M.........................................

Final................................0.25 M.......................................0.0 M................................0.25 M..........................................

By using Henderson-Hasselbalch equation :

pH = pKa + log [CH3COONa]/[CH3COOH]

pH = 4.88 + log 0.25 M / 0.25 M

pH = 4.88 + log 1

pH = 4.88 + 0.0

pH = 4.88

So, pH after the addition of 0.05 M NaOH becomes = 4.88

---------------

(c). Addition of 0.05M HCl:-

ICF table is :

....................................CH3COONa (aq)..........+.............HCl (aq) --------------> CH3COOH (aq)...........+.............NaCl (aq)

Initial ..............................0.20 M......................................0.05 M...............................0.30 M...............................................

Change...........................-0.05 M.....................................-0.05 M..............................+0.05 M.........................................

Final................................0.15 M.......................................0.0 M................................0.35 M..........................................

By using Henderson-Hasselbalch equation :

pH = pKa + log [CH3COONa]/[CH3COOH]

pH = 4.88 + log 0.15 M / 0.35 M

pH = 4.88 - 0.368

pH = 4.51

So, pH after the addition of 0.05 M HCl becomes = 4.51
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