What is the concentration (m) of sodium ions in 4.57 l of a .398 m na3p solution?

 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

b) nuclear ive learned that in a pedtrodic table