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.
mol AgNO3 = M x L
= 0.45 M x 0.045 M
mol NaCl = M x L
= 0.0135 M x 0.085 L
(Ag^+) = mols / total volume
= 0.02025 / 0.130 L
total volume = 85 mL + 45 mL
= 130 mL
= 0.130 L
(Cl^-) = mols / total volume
= 0.0011475 / 0.130 L
Ion product = [Ag^+][Cl^-]
Is my answer correct?
- chemistry - Abigail, Wednesday, April 16, 2008 at 9:13pm
b) Would a precipitate form? The Ksp of AgCl(s) is 1.8 x 10-10.
Also, please check if my answer to be is also correct or not
When comparing the ion product to Ksp, I see that 0.001375 is greater than 1.8 x 10^-10, therefore a precipitate will form.
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