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.
Answer This Question
More Related Questions
- chemistry - Question:A solution of AgNO3 (45 mL/0.45 M) was mixed with solution ...
- chemistry - Please judge my answer Question:A solution of AgNO3 (45 mL/0.45 M) ...
- chemistry - An AgNO3 solution (44 mL/0.44 M) has been mixed with an NaCl ...
- Chemistry - If 845mL of a 2.5x10^-5 mol/L solution of Ni(NO3)2 was mixed with ...
- chemistry - We are doing a lab that requires finding the molarity of an unknown ...
- chemistry - We add excess NaCl solution (58.44 g/mol) to 64 mL of a solution of ...
- Cemistry - A solution is 36% silver nitrate (AgNO3) by mass. The density of this...
- chemistry - We add excess NaCl solution (58.44 g/mol) to 56 mL of a solution of ...
- Chemistry - Consider the following reaction AgNO3 (aq) + NaCl (aq) → AgCl...
- chemistry - Silver nitrate (AgNO3) reacts with sodium chloride as indicated by ...