Chemistry: BUFFERS?
A buffer that contains .11 M HY (conjugate acid) and .220 M Y- (conjugate base) has a pH of 8.77. What is the pH after .001 mol Ba(OH)2 is added to .75L of solution?
[[HY and Y- are standard forms of a conjugate acid/base pair, its not a specific buffer]].
Answer:
We can use Handerson Hasselbalch' s equation to get Ka
pH = pKa + log [ Y-] / [ HY]
8.77 = pKa + log 0.220 / 0.11
8.77 = pKa + 0.301
8.47 = pKa
Ka = 3 x 10^-9
Initial buffer has :
0.75 L x 0.11 mol/L = 0.0825 moles HY
0.75 L x 0.220 = 0.165 moles Y-
Ba(OH)2 >> Ba2+ + 2 OH-
moles OH- added = 0.001 x 2 = 0.002
The effect of the added OH- is to convert HY in Y- via the reaction
OH- + HY >> Y- + H2O
Moles HY = 0.0825 - 0.002 = 0.0805
[ HY ] = 0.0805 / 0.75 = 0.107 M
Moles Y- = 0.165 + 0.002 = 0.167
[ Y- ] = 0.167 / 0.75 = 0.223 M
pH = pK + log [ Y-] / [ HY ] = 8.47 + log 0.223 / 0.107 = 8.79
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