The concentration of sodium ion is 0.2228 M
<u><em> calculation</em></u>
find the molarity of Na₂S solution using M₁V₁=M₂V₂ formula
where M₁ = 80.0 ml
V₁= 0.174 M
M₂ = ?
V₂ = 125 ml
make M₂ the subject of the formula by diving both side by V₂
M₂= M₁V₁/V₂
M₂ = [ (80.0 ml x 0.174 M) / 125 ml]= 0.1114 M
write the equation for dissociation of Na₂S
Na₂S→ 2Na⁺(aq) + s²⁻ (aq)
from equation above the mole ratio of Na₂s : Na⁺ is 1:2 therefore the concentration of
Na⁺= 0.1114 x 2 =0.2228 M
