[nlb149]

We have 11.18mL of 3.7M NaOH and 19.11 mL of 2.7M NaCl. 
NaOH + HCl --> NaCl + H20

a) How many grams of NaCl is formed?
b) How many ppm Na+ is there after the reaction is complete?

I think part a is easy enough just use stoichiometry to find the limiting reagent and using that get the mass of the NaCl formed.  I got 20.46g doing that.
However, I'm not sure as to how to go about doing part b.  Any suggestions?

[Sam (NG)]

I guess that should be 19.11 ml of HCl.
20.46 g is incorrect. However, your method is correct.  Determine which reagent is limiting by calculating the number of moles of each. Then, considering the stoichiometry of the equation, you know the number of moles of the product.

Parts per million is a dimensionless unit...cf. percent=parts per hundred.  If you have the mass of sodium chloride, what other quantity do you need to form this unit?

lots of concentration help here: Chembuddy

[nlb149]

Yeah that should be 19.11 ml of 2.7M HCl my bad. 

I used the wrong molar mass for NaCl so I used the right mass and got 2.42g of NaCl made.

I still have no idea what I'm suppose to do for part b.

[Borek]

How much of Na⁺ together in the final mixture? In what volume?