posted by tina .
You have 3.00 L of a 2.37 M solution of NaCl(aq) called solution A. You also have 2.00 L of a 2.00 M solution 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.
The sodium ion is a spectator ion in this reaction, i.e. it does not take part. Thus we can ignore the chemistry.
Number of moles at start
3.00 L x 2.37 mole L^-1
Because 1 mole of NaCl gives 1 mole of Na+ in solution.
volume at end = 3.00 L + 2.00 L
so concentration of Na+ at end is
=3.00 L x 2.37 mole L^-1/5.00 L