Question

An `18*mL` volume of `0.200*mol*L^-1` `Na_2SO_4` is mixed with a `15*mL` volume of `0.150*mol*L^-1` `NaCl`. What is the new concentration with respect to `[Na^+]`?

Answer 1

We use the relationship `"concentration"="number of moles"/"volume of solution"`, and get......

Explanation:

We assume that the volumes are additive, and we work out the molar concentration of the new solution.

And thus.......`[Na^+]`.........

`=(18.0xx10^-3*Lxx0.20*(mol)/L+15xx10^-3*Lxx0.150*(mol)/L)/((15.0+18.0)xx10^-3L)`

`=0.177*mol*L^-1`............

What are `[Cl^-]` and `[SO_4^(2-)]`....?

Answer 2

`["Na"^"+"] = "0.286 mol/L"`

Explanation:

Step 1. Calculate the total moles of `"Na"^"+"`

(a) Moles of `"Na"^"+"` in `"Na"_2"SO"_4`

`"Moles of Na"^"+" = 0.0180 color(red)(cancel(color(black)("L Na"_2"SO"_4))) × (0.200 color(red)(cancel(color(black)("mol Na"_2"SO"_4))))/(1 color(red)(cancel(color(black)("L Na"_2"SO"_4)))) × "2 mol Na"^"+"/(1 color(red)(cancel(color(black)("mol Na"_2"SO"_4)))) = "0.007 20 mol Na"^"+"`

(b) Moles of `"Na"^"+"` in `"NaCl"`

`"Moles of Na"^"+" = 0.0150 color(red)(cancel(color(black)("L NaCl"))) × (0.150 color(red)(cancel(color(black)("mol NaCl"))))/(1 color(red)(cancel(color(black)("L NaCl")))) × "1 mol Na"^"+"/(1 color(red)(cancel(color(black)("mol NaCl")))) = "0.002 25 mol Na"^"+"`

(c) Total moles of `"Na"^"+"`

`"Total moles" = "0.007 20 mol + 0.002 25 mol" = "0.009 45 mol"`

Step 2. Calculate the total volume of the solution

`"Total volume" = "18.0 mL + 15.0 mL" = "33.0 mL" = "0.0330 L"`

Step 3. Calculate the total concentration of `"Na"^"+"`

`["Na"^"+"] = "0.009 45 mol"/"0.0330 L" = "0.286 mol/L"`