Question

Suppose

1.07g

of zinc nitrate is dissolved in

100.mL

of a

54.0mM

aqueous solution of ammonium sulfate.

Calculate the final molarity of nitrate anion in the solution. You can assume the volume of the solution doesn't change when the zinc nitrate is dissolved in it.

Be sure your answer has the correct number of significant digits.

Answer 1

Sol :-

The balanced chemical reaction is :

Zn(NO₃)₂ (aq) + (NH₄)₂SO₄ (aq) ---------------> ZnSO₄ (s) + 2 NH₄NO₃ (aq)

No. of moles of Zn(NO₃)₂ = Mass/Molar mass = 1.07 g / 189.36 g/mol = 0.00565 mol

No. of moles of (NH₄)₂SO₄ = Molarity x Volume = 0.054 M x 0.100 L = 0.0054 mol

ICE table is :

................Zn(NO₃)₂ (aq)........... +............... (NH₄)₂SO₄ (aq) ---------------> ZnSO₄ (s) .....+........... 2 NH₄NO₃ (aq)

Initial (I)...0.00565 mol...................................0.0054 mol............................0.0 mol............................0.0 mol

Change (C) ...-0.0054 mol.............................-0.0054 mol ...........................+ 0.0054 mol.................0.0108 mol

Final (F)........0.00025 mol...............................0.0 mol..................................0.0054 mol..................0.0108 mol

So,

No. of moles of .Zn(NO₃)₂ left = 0.00025 mol and

No. of moles of NO₃^- in .Zn(NO₃)₂ = 0.00050 mol

No. of moles of NO₃^- in NH₄NO₃ = 0.0108 mol

Total moles of NO₃^- = 0.0005 + 0.0108 mol = 0.0113 mol

Final Concentration of NO₃^- = No. of moles / Volume in L = 0.0113 mol/0.100 L = 0.113 M

Hence, final molarity of nitrate anion in the solution = 0.113 M