Question

200mL of 0.1 M AgNO3 is mixed with 300 mL of 0.2 M nacl . calculate the concentration of cl- ions in the resulting solution. ( ppt is formed )​

Answer 1

Answer:

0.16 is correct answer

Answer 2

The concentration of Cl- in the resulting solution is equal to 0.08 M.

• The number of moles of NaCl is equal to 0.06 (0.3*0.2)
• The number of moles of AgNO3 is equal to 0.02 (0.2*0.1)
• One mole of AgNO3 reacts with one mole of NaCl to precipitate out one mole of AgCl.
• AgNO3 is the limiting reagent, so number of moles of AgCl precipitated is equal to 0.02, which implies that 0.02 moles of NaCl is used up.
• Mole of NaCl remaining is equal to 0.04 (0.06-0.02)
• Total volume of the solution is equal to 500 mL.
• Hence molarity of NaCl, which will be equal to the molarity of Cl- ions ia equal to 0.08 M (0.04*1000/500)