Calculate the concentration (m) of sodium ions in a solution made by diluting 80.0 ml of a 0.174 m solution of sodium sulfide to a total volume of 125 ml.

The concentration of sodium ion is 0.2228 M

calculation

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