# 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^+]?

Aug 5, 2017

We use the relationship $\text{concentration"="number of moles"/"volume of solution}$, and get......

#### Explanation:

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

And thus.......$\left[N {a}^{+}\right]$.........

$= \frac{18.0 \times {10}^{-} 3 \cdot L \times 0.20 \cdot \frac{m o l}{L} + 15 \times {10}^{-} 3 \cdot L \times 0.150 \cdot \frac{m o l}{L}}{\left(15.0 + 18.0\right) \times {10}^{-} 3 L}$

$= 0.177 \cdot m o l \cdot {L}^{-} 1$............

What are $\left[C {l}^{-}\right]$ and $\left[S {O}_{4}^{2 -}\right]$....?

Aug 6, 2017

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

#### Explanation:

Step 1. Calculate the total moles of $\text{Na"^"+}$

(a) Moles of $\text{Na"^"+}$ in ${\text{Na"_2"SO}}_{4}$

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

(b) Moles of $\text{Na"^"+}$ in $\text{NaCl}$

$\text{Moles of Na"^"+" = 0.0150 color(red)(cancel(color(black)("L NaCl"))) × (0.150 color(red)(cancel(color(black)("mol NaCl"))))/(1 color(red)(cancel(color(black)("L NaCl")))) × "1 mol Na"^"+"/(1 color(red)(cancel(color(black)("mol NaCl")))) = "0.002 25 mol Na"^"+}$

(c) Total moles of $\text{Na"^"+}$

$\text{Total moles" = "0.007 20 mol + 0.002 25 mol" = "0.009 45 mol}$

Step 2. Calculate the total volume of the solution

$\text{Total volume" = "18.0 mL + 15.0 mL" = "33.0 mL" = "0.0330 L}$

Step 3. Calculate the total concentration of $\text{Na"^"+}$

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