Explanation Given The solution consists of 0.050 M Mg 2 + ( a q ) and Cu 2 + ( a q ) . The K s p value for Mg ( OH ) 2 is 1.8 × 10 − 11 and that for Cu ( OH ) 2 is 4.8 × 10 − 20 . The highest concentration of OH − that can be present without precipitation of Mg ( OH ) 2 is calculated by the formula, K s p 1 = [ Mg 2 + ] [ OH − ] 1 2 Where, K s p 1 is the K s p for Mg ( OH ) 2 . [ Mg 2 + ] is the concentration of Mg 2 + . [ OH − ] 1 is the highest concentration of OH − in Mg ( OH ) 2 . Substitute the values of K s p 1 and [ Mg 2 + ] in the above formula. 1.8 × 10 − 11 = ( 0.050 ) [ OH − ] 2 [ OH − ] 1 2 = 3.6 × 10 − 10 [ OH − ] 1 = 1

Chemistry: The Central Science (13th Edition)

13th Edition

ISBN: 9780321910417

Author: Theodore E. Brown, H. Eugene LeMay, Bruce E. Bursten, Catherine Murphy, Patrick Woodward, Matthew E. Stoltzfus

Publisher: PEARSON

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Chapter 3.3, Problem 3.6.2PE

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To determine: The ion that precipitates first as OH−(aq) is added and the concentration of OH−(aq) that is necessary to begin the precipitation of each cation in the given solution.

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Chemistry: The Central Science (13th Edition)

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The ion that precipitates first as OH − ( a q ) is added and the concentration of OH − ( a q ) that is necessary to begin the precipitation of each cation in the given solution.

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To determine: The ion that precipitates first as OH−(aq) is added and the concentration of OH−(aq) that is necessary to begin the precipitation of each cation in the given solution.

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Chapter 3 Solutions