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March 29, 2015

Homework Help: chemistry

Posted by Abigail on Wednesday, April 16, 2008 at 9:09pm.

Question:A solution of AgNO3 (45 mL/0.45 M) was mixed with solution of NaCl (85 mL/1.35 x 10-2 M)
a) Calculate the ion product of the potential precipitate.


My answer:

mol AgNO3 = M x L
= 0.45 M x 0.045 M
= 0.02025

mol NaCl = M x L
= 0.0135 M x 0.085 L
= 0.0011475

(Ag^+) = mols / total volume
= 0.02025 / 0.130 L
= 0.1558


total volume = 85 mL + 45 mL
= 130 mL
= 0.130 L


(Cl^-) = mols / total volume
= 0.0011475 / 0.130 L
= 0.00883

Ion product = [Ag^+][Cl^-]
= [0.1558][0.00883]
= 0.001375

Is my answer correct?

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