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

answer

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