What's the M of the NaOH after mixing and before reaction?
0.200 M NaOH x (14.0/(14.0+42.4) = approx 0.00025. Since there is 1 Na^+ in 1 molecule of NaOH, 0.00025 is the M of the Na^+.
The nitrate is done the same way. You need not worry about the reaction since Na^+ and NO3^- are spectator ions. By the way that 0.00025 is an estimate; you should recalculate more accurately.
A solution of 0.200M NaOH (14.0 mL) is mixed with 42.4 mL of 0.170M HNO3. Assuming the final volume is the sum of the initial volumes, what is the molarity of Na+ cation and NO3- anion?
I do not know where to start to solved for the molarity of the ions. Please help !
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