HW3_10

# HW3_10 - 2 H 3 O 2 2 Ca(C 2 H 3 O 2 2> Ca 2 2 C 2 H 3 O 2...

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Homework due 3/10 Calculate the pH of a solution that is 0.50 M in HC 2 H 3 O 2 and 0.30 M in Ca(C 2 H 3 O 2 ) 2 . K a for HC 2 H 3 O 2 = 1.8x10 -5 This is a typical common ion problem in which the solution contains a weak acid and it salt. The initial concentration of the conjugate base of the acid is determined from the concentration of the salt added, assuming complete dissociation. The acid dissociation expression can then be used to determine the [H 1+ ], hence the pH of the solution at equilibrium. NOTE: since Ca(C 2 H 3 O 2 ) 2 dissociates to give 2 moles C 2 H 3 O 2 1- ion, its concentration is double that of the original Ca(C
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Unformatted text preview: 2 H 3 O 2 ) 2 Ca(C 2 H 3 O 2 ) 2--------------> Ca 2+ + 2 C 2 H 3 O 2 1-0.30 M 0.30 M 0.60 M HC 2 H 3 O 2 H 1+ + C 2 H 3 O 2 1-K a = [H 1+ ][C 2 H 3 O 2 1-] / [HC 2 H 3 O 2 ] [HC 2 H 3 O 2 ] [H 1+ ] [C 2 H 3 O 2 1-] initial 0.50 0.60 change- x x x equilibrium 0.50 – x x 0.60 + x where x = [H 1+ ] produced K a = [H 1+ ][C 2 H 3 O 2 1-] / HC 2 H 3 O 2 ] = (x)(0.60 + x) / (0.50 – x) = 1.8x10-5 ignoring the added and subtracted x’s gives (x) (0.60) / (.50) = 1.8x10-5 x = [H 1+ ] = 1.5x10-5 pH = 4.82...
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