A 1.40 g sample of silver nitrate is dissolved in sufficient water to make 125 mL of solution. To this solution is added 5.00 mL of 1.50 M hydrochloric acid, and a precipitate forms. Find the concentration of silver ions remaining in solution.
I have scoured the internet and not found a question similar to this. I can't figure out what to do. I know that the AgNO3 + HCL -> AgCl + HNO3 and that tthe AgCl will be a precipitate. From this, I'm supposed to find out how much silver is left in the solution. I just cannot piece together how all the information provided will get me to the final answer. If you are able to only provide the steps I'm supposed to do and not the actual math, that would still be great. Thanks in advance.
I have scoured the internet and not found a question similar to this. I can't figure out what to do. I know that the AgNO3 + HCL -> AgCl + HNO3 and that tthe AgCl will be a precipitate. From this, I'm supposed to find out how much silver is left in the solution. I just cannot piece together how all the information provided will get me to the final answer. If you are able to only provide the steps I'm supposed to do and not the actual math, that would still be great. Thanks in advance.
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AgNO3 + HCl → AgCl + HNO3
(1.40 g AgNO3) / (169.8732 g AgNO3/mol) = 0.0082414 mol AgNO3
(0.00500 L) x (1.50 mol HCl/L) = 0.0075 mol HCl = 0.0075 mol AgNO3 reacted
(0.0082414 mol initially) - (0.0075 mol reacted) = 0.000741 mol AgNO3 unreacted
Supposing the volumes to be additive: 125 mL + 5.00 mL = 130 mL
final concentration:
(0.000741 mol AgNO3) / (0.130 L) = 0.00570 mol/L AgNO3 = 0.00570 mol/L Ag{+}
(1.40 g AgNO3) / (169.8732 g AgNO3/mol) = 0.0082414 mol AgNO3
(0.00500 L) x (1.50 mol HCl/L) = 0.0075 mol HCl = 0.0075 mol AgNO3 reacted
(0.0082414 mol initially) - (0.0075 mol reacted) = 0.000741 mol AgNO3 unreacted
Supposing the volumes to be additive: 125 mL + 5.00 mL = 130 mL
final concentration:
(0.000741 mol AgNO3) / (0.130 L) = 0.00570 mol/L AgNO3 = 0.00570 mol/L Ag{+}