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^+]#?

2 Answers
Aug 5, 2017

Answer:

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-)]#....?

Aug 6, 2017

Answer:

#["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"#