A 40.0-mL solution of 0.040 0 M Hg 2 (NO 3 ) 2 was titrated with 60.0
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A 40.0-mL solution of 0.040 0 M Hg2(NO3)2 was titrated with 60.0 mL of 0.100 M KI to precipitate Hg2I2 (Ksp = 4.6 × 10-29).
(a) Show that 32.0 mL of KI are needed to reach the equivalence point.
(b) When 60.0 mL of KI have been added, virtually all Hg22+ has precipitated, along with 3.20 mmol of I-. Considering all ions remaining in the solution, calculate the ionic strength when 60.0 mL of KI have been added.
(c) Using activities, calculate pHg2+2 (= -log AHg22+) for part (b).
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a 40 mL 004 M Hg 2 NO 3 2 16 mmol of Hg 2 2 ions or 32 mmol of Hg 2 ion Hence 32mL 01M KI s...View the full answer
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