Asked by Anonymous
When 50 mL of 1.0 M AgNO3 is added to 50 mL of 0.50 M HCl, a precipitate of AgCl forms. After the reaction is complete, what is the concentration of silver ions in the resulting solution?
(a) 0.50 M
(b) 1.0 M
(c) zero
(d) 0.25 M
(e) 0.75 M
How do I go about solving this?
I attempted to start it.. I'm looking to find concentration, correct? So I would use molarity, which = moles solute / liters solution
M= mol solute / L soln
How do I find the moles of solute? which is the solute, is it AgNO3?
Would the L of soln be 0.1 L, because 50 mL + 50 mL = 100 mL = 0.1 L?
Please help!
(a) 0.50 M
(b) 1.0 M
(c) zero
(d) 0.25 M
(e) 0.75 M
How do I go about solving this?
I attempted to start it.. I'm looking to find concentration, correct? So I would use molarity, which = moles solute / liters solution
M= mol solute / L soln
How do I find the moles of solute? which is the solute, is it AgNO3?
Would the L of soln be 0.1 L, because 50 mL + 50 mL = 100 mL = 0.1 L?
Please help!
Answers
Answered by
DrBob222
Yes, the volume will be 0.05L + 0.05L= 0.1 L.
If you started with 0.05 L of 1 M AgNO3, you had 0.05 moles Ag ion initially.
You added 0.05 L x 0.5 M = 0.025 moles HCl.
So 0.025 moles Ag ion were used leaving
0.050-0.025 = 0.025 moles Ag ion (and I'm assuming we neglect the VERY small amount of Ag ion that comes from the AgCl ppt).
M = moles/L. 0.025/0.1 =
If you started with 0.05 L of 1 M AgNO3, you had 0.05 moles Ag ion initially.
You added 0.05 L x 0.5 M = 0.025 moles HCl.
So 0.025 moles Ag ion were used leaving
0.050-0.025 = 0.025 moles Ag ion (and I'm assuming we neglect the VERY small amount of Ag ion that comes from the AgCl ppt).
M = moles/L. 0.025/0.1 =
Answered by
Anonymous
Thank you!
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