100 mL buffer x 0.1M CH3COOH = 10 mmoles.
100 mL buffer x 0.1M CH3COONa = 10 mmoles.
add 1 mL x 0.1M HCl = 0.1mmole.
Let's call acetic acid, CH3COOH, HAc.
...........Ac^- + H^+ ==> HAc
initial.....10.....0.......10
add...............0.1..............
change.....-0.1..-0.1.....+0.1
equil.......9.9.....0.....10.1
Plug into the Henderson-Hasselbalch equation and solve for pH. Using 1.75E-5 for Ka I obtained a little over 4.7 but you need to do it more accurately AND your text/notes may list a different value for Ka.
Calculate z pH of a buffer solun containing o.1 M acetic acid & 0.1 M solun of sodiumacetate when 1 mL of 0.1M HCl is added to 100 mL of buffer .
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