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Horvath Practice Problems 59

# Horvath Practice Problems 59 - HC 6 H 4 NO 2 NH 3 ­ C 6 H...

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From this K b 2 = 6 . 3 × 10 3 = ( x ) ( x ) (0 . 05 x ) Rearranging and using the quadratic formula we get x = £ OH ¤ = 0 . 015 and pOH = log (0 . 015) = 1 . 82 . So, fi nally, pH = 14 1 . 82 = 12 . 18 Reaction Between a Weak Acid and a Weak Base The reaction between a weak acid and a weak base requires a di ff erent strategy since we can not exploit the full dissociation of the strong acid or base as we did above. We can perform this calculation by using Hess’ law from last semester. Example : What are the fi nal concentrations of products if 25 ml of a 0.10 M niacin solution ( K a = 1 . 4 × 10 5 ) is reacted with 25 ml of a 0.10 M NH 3 solution ( K b = 1 . 8 × 10 5 )? As before, we start with a balanced equation
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Unformatted text preview: HC 6 H 4 NO 2 + NH 3 ­ C 6 H 4 NO − 2 + NH + 4 . We now use Hess’ law to write HC 6 H 4 NO 2 ­ H + + C 6 H 4 NO − 2 K a = 1 . 4 × 10 − 5 NH 3 + H 2 O ­ NH + 4 + OH − K b = 1 . 8 × 10 − 5 H + + OH − ­ H 2 O 1 /K w = 1 . × 10 +14 This gives the original reaction and the equilibrium constant for this reaction is K = K a K b K w = 2 . 5 × 10 +4 Upon mixing the volume of the solution is doubled so the initial concentrations are half of what they are prior to mixing. 58 REACTIONS WITH WEAK ACIDS AND WEAK BASES...
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