An 18*mL18mL volume of 0.200*mol*L^-10.200molL1 Na_2SO_4Na2SO4 is mixed with a 15*mL15mL volume of 0.150*mol*L^-10.150molL1 NaClNaCl. What is the new concentration with respect to [Na^+][Na+]?

2 Answers
Aug 5, 2017

We use the relationship "concentration"="number of moles"/"volume of solution"concentration=number of molesvolume 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^+][Na+].........

=(18.0xx10^-3*Lxx0.20*(mol)/L+15xx10^-3*Lxx0.150*(mol)/L)/((15.0+18.0)xx10^-3L)=18.0×103L×0.20molL+15×103L×0.150molL(15.0+18.0)×103L

=0.177*mol*L^-1=0.177molL1............

What are [Cl^-][Cl] and [SO_4^(2-)][SO24]....?

Aug 6, 2017

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

Explanation:

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

(a) Moles of "Na"^"+"Na+ in "Na"_2"SO"_4Na2SO4

"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"