Question

What is the concentration (M) of sodium ions in 4.57 L of a .398 M Na3P solution?

Answer 1

 The   concentration   of sodium  ion  is  1.194 M

 calculation

 Step 1:  write  the equation for  dissociation  of Na₃P

Na₃P  →  3 Na⁺(aq)  + P⁻ (aq)

  Step  2:  find the  moles of  Na₃p

MOLES =  molarity  x volume in liters

 =  4.57 L  x  0.398  M = 1.819  moles

Step 3:  use  the  mole ratio to calculate the  moles of  Na⁺

Na₃P: Na⁺  is   1:3  therefore the  moles of Na⁺  =   3  x 1.819  = 5.457  moles

Step  4 :  find  the concentration (molarity)  of Na⁺

molarity  =  moles /volume in liters

=5.457  moles / 4.57 = 1.194 M

Answer 2

Answer:

The concentration (M) of sodium ions in 4.75 L solution is 1.19 mol/L.

Explanation:

Molarity : It is defined as the number of moles of solute present in one liter of solution.  Mathematically written as:

$Molarity=\frac{\text{Moles of solute}}{\text{volume of solution in L}}$

Moles of sodium phosphide = n

Volume of the solution = 4.57 L

Molarity of the solution = $Na_3P]$ = 0.398 M

$0.398 M=\frac{n}{4.57 L}$

n  = 0.398 M × 4.57 L = 1.819 moles

$Na_3P(aq)\rightarrow Na^+(aq)+P^{3-}(aq)$

1 mole gives sodium phosphide gives 3 moles of sodium ions and 1 mole of phosphide ions.

Then 1.819 moles of sodium phosphide will give :

$3\times 1.819 mol=5.457 mol$ of sodium ions.

The concentration (M) of sodium ions in 4.75 L solution is:

$[Na^+]=\frac{5.457 mol}{4.57 L}=1.194 mol/L\approx 1.19 mol/L$