4) Calculate the mass of AgCl formed, and the concentration of silver ion remaining in solution, when 10.0g of solid AgNO3 is added to 50mL of 1.00 X 10^-2 M NaCl solution. Assume the volumes are additive.
I got the mass of AgCl to be .0717g but Could you tell me how to find the concentration of remaining silver ion?
I got the mass of AgCl to be .0717g but Could you tell me how to find the concentration of remaining silver ion?
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AgNO3 + NaCl → AgCl + NaNO3
(0.050 L) x (1.00 x 10^-2 mol/L) x (1/1) x (143.3214 g/mol) = 0.0717 g AgCl
So you are correct about the mass of AgCl.
(10.0 g AgNO3) / (169.8732 g AgNO3/mol) = 0.0589 moles AgNO3 at start
(0.050 L) x (1.00 x 10^-2 mol/L) x (1/1) = 0.0005 mole AgNO3 used
0.0589 moles - 0.0005 mole = 0.0584 mole AgNO3 (and Ag+ ions) remaining
concentration = (0.0584 mole / 0.050 L) = 1.17 M of Ag+ ions remaining
(But to do this really right, you'd need to find the volume of 10.0 g of AgNO3 then add it to the 50 mL of original solution in order to "assume the volumes are additive", but I think they don't really mean it, since volumes are decidedly not additive when dealing with a solid and a liquid.)
(0.050 L) x (1.00 x 10^-2 mol/L) x (1/1) x (143.3214 g/mol) = 0.0717 g AgCl
So you are correct about the mass of AgCl.
(10.0 g AgNO3) / (169.8732 g AgNO3/mol) = 0.0589 moles AgNO3 at start
(0.050 L) x (1.00 x 10^-2 mol/L) x (1/1) = 0.0005 mole AgNO3 used
0.0589 moles - 0.0005 mole = 0.0584 mole AgNO3 (and Ag+ ions) remaining
concentration = (0.0584 mole / 0.050 L) = 1.17 M of Ag+ ions remaining
(But to do this really right, you'd need to find the volume of 10.0 g of AgNO3 then add it to the 50 mL of original solution in order to "assume the volumes are additive", but I think they don't really mean it, since volumes are decidedly not additive when dealing with a solid and a liquid.)