In the solution containing both 0.10 M acetic acid and 0.10 M sodium acetate, the acetic...

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In the solution containing both 0.10 M acetic acid and 0.10 Msodium acetate, the acetic acid undergoes ionization. The chemicalequation for this ionization reaction is the same as for a solutioncontaining acetic acid alone. The difference is that the initialconcentration of acetate ion (before any ionization reactionoccurs) for the solution containing acetic acid alone is zero,whereas the initial concentration of acetate ion is 0.10 M in yoursolution containing both acetic acid and sodium acetate. Calculatethe percent ionization and the expected initial pH for the solutionthat contained both 0.10 M acetic acid and 0.10 M sodium acetate.(Hint: Again, you will need to use Ka for acetic acid.) PercentIonization Calculated pH

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pka of aceticacid = -logka

Â Â Â Â Â Â Â Â Â Â Â Â = -log(1.75*10^âˆ’5)
Â Â Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Â Â Â Â = 4.76

aceticacid + CH3COONa = buffer

pH of acidic buffer = pka + log(salt/acid)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 4.76 + log(0.1/0.1)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 4.76

Â Â Â Â Â Â Â Â Â Â Â Â pH = 4.76

Â Â Â Â Â Â Â Â Â Â CH3COOH (aq) <-----> CH3COO-(aq) + H^+(aq)

Â Â initial 0.1 MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.1 MÂ Â Â Â Â Â 0

Â Â changeÂ Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â x

equilibrium 0.1-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â 0.1+xÂ Â Â Â Â Â Â x

Â Â Â ka = [CH3COO-][H+]/[CH3COOH]

Â Â Â 1.75*10^-5 = ((0.1+X)X)/(0.1-X)

Â Â Â X = 1.75*10^-5 M

percentage dissociation of CH3COOH = dissociated / initial*100

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (1.75*10^-5)/0.1*100

Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.0175 %

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In the solution containing both 0.10 M acetic acid and 0.10 M sodium acetate, the acetic...

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Chemistry

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