## Answers

page-01 Answers (©) when 35 ml. d 0.02570 M CalNog), added 2- 25.00 ml Solution contains = 0.03110M (N9, 9204) it is breated with Ca(NO3)2 = 0.02570 M given geaction : Calt +6,042-cac204 (5) after equivalence point, there is excess of ca²+ aons moles of cazt = added moles de capta moles of czby added (0.0254) (0.02570 molle) -(0.025%)(0,03107 = 0.0008995 mol -0.0007775 med. =1.22x10-4 moles of cakt = 0.000122 mol volume of solution after addition 85 ml of 0.02570 M Ca (NO), (0.025 +0.035) = 0.06

page-02 [cast] = 0.000192 met 0.06L = 2.033 X 107 = 0.002033 M p Ca²+ = -log[cat] = -log (0.002033] = 2.6918 1pcal+ = 2.69

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