Explanation The overall idea is to calculate the equilibrium constants given the concentrations and pH data. Note that this requires the relationship between species in equilibrium. Let the weak base be characterized as “B” and BH+ as the hydrolysed base. Step 1: Hydrolysis of morphine B ( a q ) + H 2 O ( l ) → B H + ( a q ) + O H − ( a q ) Step 2: the equilibrium expression for the weak base K b = [ B H + ] [ O H − ] [ B ] Step 3: Using the given data, relate concentrations get “x” Initially [ B ] = 7.00 x 10 − 4 M [ B H + ] = 0 [ O H − ] = 0 Let “x” be the dissociated part of the base. In equilibrium: [ B ] = 7.00 x 10 − 4 − x [ B H + ] = 0 + x [ O H − ] = 0 + x From pH data, get pOH, then hydroxide concentration p H + p O H = 14 p O H = 14 − p H p O H = 14 − 9.50 = 4.50 [ O H − ] = 10 − p O H = 10 − 4

Chemistry, Books a la Carte Edition and Modified Mastering Chemistry with Pearson eText & ValuePack Access Card (7th Edition)

7th Edition

ISBN: 9780134172514

Author: John E. McMurry

Publisher: PEARSON

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Chapter 15, Problem 15.106SP

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The constants and power constant for morphine.

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Chapter 15, Problem 15.105SP

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Chemistry, Books a la Carte Edition and Modified Mastering Chemistry with Pearson eText & ValuePack Access Card (7th Edition)

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To determine:

The constants and power constant for morphine.