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What is the pH at the equivalence point when 55.0 mL of a 0.300 M solution...

What is the pH at the equivalence point when 55.0 mL of a 0.300 M solution of acetic acid (CH3COOH) is titrated with 0.100 M NaOH to its end point?
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Answer #1

Since this isia weak acid- base titration so,

CH3COOH having 55.0ml(. 055 l) and .3M

for mole, . 055* . 3=0.0165 mole

So for NaOH 0.1650 l mixed with 0.1M soln to have 0.0165 mole of NaOH

NaOH + CH3CHOOH = NaCH3COO + H2O

Initial 0.0165 0.0165 0

Equilibrium 0 0 0.0165

Now Molarity of NaCH3COO = mole/litre

= 0.076

Now,

NaCH3COO + H2O = NaCH3COOH + OH-

Initial 0.075 0 0

Eqm 0.075-x x x

Kb = [NaCH3COOH] [OH-] /[NaCH3COO]

Kw=Kb*ka here ka for given acid is = 1.8*10^-10 and kw= 10^-14 so kb= 5.6*10^-10

Therefore, 5.6*10^-10 = x^2/0.075 since 0.075-x = 0.075

So x= 6.48*10^-6

Now pOH = - Log[6.48*10^-6]

= 5.188

Therefore pH = 14-5.188

= 8.812 answer

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