Ans 1
[OH-] = 0.000045 M
pOH = - log [OH-] = - log (0.000045) = 4.3468
pH = 14 - pOH
= 14 - 4.3468
= 9.6532
Ka = 6*10^-9
pKa = - logKa = - log(6*10^-9) = 8.222
From the Henderson Hassalbalch equation
pH = pKa + log([acetate] / [acetic acid])
9.6532 = 8.222 + log ([acetate] / [0.74])
1.4312 = log ([acetate] / [0.74])
26.989 = [ acetate] / [0.74]
[ acetate] = 19.972 M
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