Calculate the concentration (in M) of Cl- ions in solution C

[AMan24]

Homework Statement

[/B]
PROBLEM #1 You have 3.00 L of a 3.39 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 Msolution of AgNO3(aq) called solution B. You mix these solutions together, making solution C.

Calculate the concentration (in M) of Cl- ions in solution C.PROBLEM #2 You have 3.00 L of a 3.00 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 Msolution of AgNO3(aq) called solution B. You mix these solutions together, making solution C.

Calculate the concentration (in M) of Na+ ions in solution C.

Homework Equations

nope

The Attempt at a Solution

For problem #2, i figured it out easily. I just did:
3.00M x 3.00L = 9moles.
9moles/(2.00L + 3.00L) = 1.80M

For problem #1, i found the answer but i don't understand why. Heres what its supposed to be like:

3.39M x 3.00L = 10.17moles Nacl.
2.00M x 2.00L = 4moles AgNO3 <------DON'T UNDERSTAND WHY IM FINDING BOTH FOR THIS ONE AND NOT THE OTHER

now you do 10.17 moles NaCl - 4moles AgNO3 = 6.17Moles, (no idea why I am subtracting for this and not the other).

Now divide 6.17 by the total like the other problem. 6.17/(2.00L + 3.00L) = 1.234M...

- i don't think its because one is a cation and one is an anion
- my guess is that its because AgCl is a solid and NaNO3 is aqueous

 

[DrClaude]

- my guess is that its because AgCl is a solid and NaNO3 is aqueous

Yes. Write down the balanced equation for what happens when you mix NaCl and AgNO₃, with proper labels [(aq), etc.], and see what is in solution.

 

[Borek]

For problem #2, i figured it out easily. I just did:
3.00M x 3.00L = 9moles.
9moles/(2.00L + 3.00L) = 1.80M

For the record: that's not a correct answer.