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
The concentration of [tex]\rm Na^+[/tex] in the solution is 1.194 M.
The moles of [tex]\rm Na_3P[/tex]solution:
Moles = molarity [tex]\times[/tex] volume
Moles = 0.398 [tex]\times[/tex] 4.57
moles of [tex]\rm Na_3P[/tex] = 1.819 moles.
The dissociation of [tex]\rm Na_3P[/tex] yields:
[tex]\rm Na_3P\;\rightarrow\;3\;Na^+\;+\;P[/tex]
1 mole of [tex]\rm Na_3P[/tex] yields 3 moles of [tex]\rm Na^+[/tex]
1.819 moles of [tex]\rm Na_3P[/tex] = 3 [tex]\times[/tex] 1.819 moles of [tex]\rm Na^+[/tex]
Moles of [tex]\rm Na^+[/tex]= 5.457 moles
Molarity of [tex]\rm Na^+[/tex] = [tex]\rm \frac{moles}{volume}[/tex]
Molarity of [tex]\rm Na^+[/tex]= [tex]\rm \frac{5.457}{4.57}[/tex]
Molarity of [tex]\rm Na^+[/tex]= 1.194 M.
The concentration of [tex]\rm Na^+[/tex] in the solution is 1.194 M.
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