Question

If 100. mL of 0.400 M Na2SO4 is added to 200. mL of 0.600 M NaCl, what is the concentration of Na+ ions in the final solution? Assume that the volumes are additive.

Answer 1

Na₂SO₄ → 2Na⁺ + SO₄²⁻

c₁(Na⁺)=2c(Na₂SO₄)

n₁(Na⁺)=c₁(Na⁺)v₁=2c(Na₂SO₄)v₁

NaCl → Na⁺ + Cl⁻

c₂(Na⁺)=c(NaCl)

n₂(Na⁺)=c₂(Na⁺)v₂=c(NaCl)v₂

n(Na⁺)=n₁(Na⁺)+n₂(Na⁺)=2c(Na₂SO₄)v₁+c(NaCl)v₂

v=v₁+v₂

c(Na⁺)=n(Na⁺)/v={2c(Na₂SO₄)v₁+c(NaCl)v₂}/(v₁+v₂)

c(Na⁺)={2*0.400*0.1+0.600*0.2}/(0.1+0.2)=0.667 mol/L

c(Na⁺)=0.667M

Answer 2

The concentration of Na ion in the final solution is 0.66 M.

To calculate the concentration of Na ions in the solution, the moles of $\rm Na_2SO_4$ in the solution have been calculated.

Moles = molarity $\times$ volume (L)

The molarity of $\rm Na_2SO_4$ solution given is = 0.4

Volume of $\rm Na_2SO_4$ solution = 100 ml = 0.1 L.

Moles of $\rm Na_2SO_4$ = 0.4 $\times$ 0.1

Moles of $\rm Na_2SO_4$ = 0.04

The dissociation of $\rm Na_2SO_4$ will be as follows:

$\rm Na_2SO_2\;\rightarrow\;2\;Na^+\;+\;SO_4^2^-$

1 mole of $\rm Na_2SO_4$ = 2 moles of Na ions.

0.04 moles of $\rm Na_2SO_4$ = 0.04 $\times$ 2

0.04 moles of $\rm Na_2SO_4$ = 0.08 moles of Na ions.

The moles of Na ions in NaCl solution will be:

Given, molarity of NaCl solution = 0.6 M

Volume of NaCl solution = 200 ml = 0.2 L.

Moles of Na ions in NaCl solution = 0.6 $\times$ 0.2

Moles of Na ions in NaCl solution = 0.12.

The dissociation of NaCl will be as follows:

$\rm NaCl\;\rightarrow\;Na^+\;+\;Cl^-$

1 mole of NaCl = 1 mole of Na ions

0.12 moles of NaCl gives = 0.12 moles of Na ions.

Thus the total Na ions = Na ion in $\rm Na_2SO_4$ + Na ions in NaCl

Total moles of Na ions = 0.08 + 0.12

Total moles of Na ions = 0.2 moles.

The total volume of solution will be :

0.1 L $\rm Na_2SO_4$ + 0.2 L NaCl

= 0.3 L.

Concentration of Na ion in the final solution = $\rm \dfrac{moles}{volume\;(L)}$

The concentration of Na ion in the final solution =  $\rm \dfrac{0.2}{0.3}$ M

Concentration of Na ion in the final solution = 0.66 M.

The concentration of Na ion in the final solution is 0.66 M.

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