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Calculate the pH of a solution of 0.100M HC2H3O2 (acetic acid) and 0.125M NaC2H3O2 (sodium acetate)...

Calculate the pH of a solution of 0.100M HC2H3O2 (acetic acid) and 0.125M NaC2H3O2 (sodium acetate) using an ICE (ICF) table.
Calculate the pH of a solution of 0.010M C5H5NH+Cl- (pyridinium hydrochloride) and 0.006M C5H5N (pyridine) using an ICE (ICF) table.

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Homework Answers

Answer #1

(1) pKa of acetic acid = 4.75

From Henderson equation

pH = pKa + log {[Conjugate base] / [Acid]}

pH = 4.75 + log {[NaC2H3O2] / [HC2H3O2]}

pH = 4.75 + log (0.125 / 0.100)

pH = 4.75 + log 1.25

pH = 4.75 + 0.097 = 4.85

pH = 4.85

(2) pKb of pyridine= 8.75

We know from Henderson equation

pOH = pKb + log {[Conjugate acid] / [Base]}

pOH = 8.75 + log{[pyridinium hydrochloride] / [pyridine]}

pOH = 8.75 + log(0.010 / 0.006)

pOH = 8.75 + log 1.67 = 8.75 + 0.22 = 8.97

pH = 14.00 - pOH = 14.00 - 8.97 = 5.03

pH = 5.03

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