Explanation Given Solubility product constant for PbSO 4 is 6.3 × 10 − 7 . Solubility product constant for SrSO 4 is 3.2 × 10 − 7 . The dissolution of PbSO 4 in water is given as, PbSO 4 ( s ) ⇌ Pb 2 + ( a q ) + SO 4 2 − ( a q ) The dissolution of SrSO 4 in water is given as, SrSO 4 ( s ) ⇌ Sr 2 + ( a q ) + SO 4 2 − ( a q ) The concentration of [ Pb 2 + ] is calculated by the formula, K s p 1 = [ Pb 2 + ] [ SO 4 2 − ] [ SO 4 2 − ] = K s p 1 [ Pb 2 + ] ……(1) Where, K s p 1 is the solubility product constant for PbSO 4 . [ SO 4 2 − ] is the concentration of SO 4 2 − in PbSO 4 . The concentration of [ Sr 2 + ] is calculated by the formula, K s p 2 = [ Sr 2 + ] [ SO 4 2 − ] [ SO 4 2 − ] = K s p 2 [ Sr 2 + ] ……(1) Where, K s p 2 is the solubility product constant for PbSO 4 . [ SO 4 2 − ] is the concentration of SO 4 2 − in SrSO 4 . As both the solutions are in equilibrium thus the concentration of SO 4 2 − in each of them is equal. Thus, the above two equations ate equal and the resultant formula is, K s p 1 [ Pb 2 + ] = K s p 2 [ Sr 2 + ] [ Pb 2 + ] [ Sr 2 + ] = K s p 1 K s p 2 Substitute the values of K s p 1 and K s p 2 in the above formula. [ Pb 2 + ] [ Sr 2 + ] = 6

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, Problem 85E

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To determine: The values of [SO42−] , [Pb2+] and [Sr2+] in a solution that is in equilibrium with PbSO4 and SrSO4 .

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Chapter 3, Problem 84E

Chapter 3, Problem 86AE

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Chemistry: The Central Science (13th Edition)

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The values of [ SO 4 2 − ] , [ Pb 2 + ] and [ Sr 2 + ] in a solution that is in equilibrium with PbSO 4 and SrSO 4 .

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To determine: The values of [SO42−] , [Pb2+] and [Sr2+] in a solution that is in equilibrium with PbSO4 and SrSO4 .

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Chapter 3 Solutions