What is the mass of the precipitate formed when 12.0 mL of 0.150 M NaCl is added to 25.00 mL of a 0.0500 M AgN

Question

What is the mass of the precipitate formed when 12.0 mL of 0.150 M NaCl is added to 25.00 mL of a 0.0500 M AgNO3 solution?

Answer 1

The Ksp of AgCl is very very small, that means most of the Ag and Cl will percipiate, so we can ignore solubility products. Therefore:

(.0500 M Ag+)(.025 L) =.00125 moles Ag+

all (the Ksp is too small) the Ag+ will percipitate to form AgCl, so .00125 moles AgCl

(.00125 m AgCl)(143.29 g/m)=.1791 grams percipitate

Answer 2

you'll choose the solubility consistent for BaCO3 (called Ksp) for barium carbonate, the ionization reaction will be: BaCO3 ---> Ba2+ + CO3(2-) by skill of definition, the fee of Ksp would have the formulation: Ksp = [Ba2+][CO3(2-)] the position the brackets factor out molar concentrations of the stated ions Ksp are frequently present day in literature and given in the time of checks. you could ascertain this concentrations from the given. if the made from the most concentration is below Ksp, no precipitation will happen. in the different case, BaCO3 will precipitate out of the answer.

Answer 3

You first need to balance the equation.
2 NaCl + AgN = 2 Na+ + N2- + AgCl2 (solid)

Next convert the ml to mmol using the Molarity.
for example 12 ml NaCl X 0.150 Mol/ml = 1.8 mmol NaCl
25 ml AgN X .05 Mol/ml = 1.25 mmol AgN

Use these numbers with the stoichiometric coefficients from the balanced equation to find that we have 1.25 mmol of the precipitate, AgCl2.
Example: 1.25 mmol AgN is proportional to X mmols AgCl2. Since the ratio from the balanced equation is 1:1 we know there are 1.25 mmol of AgCl2. Then we just use the molecular weight of AgCl2 (109 +71 = 180 mg/mmol) than use unit analysis to find the mg of precipitate.
1.25 mmol AgCl2 X 180 mg/mmol = 225 mg AgCl2 Your Answer!

Answer 4

What Is The Precipitate