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What concentration of [tex]\text{ClO}_3^-[/tex] results when 821 mL of 0.349 M [tex]\text{AgClO}_3[/tex] is mixed with 913 mL of 0.267 M [tex]\text{Mn(ClO}_3)_2[/tex]?

Answer :

The concentration of ClO₃⁻ in the resulting solution is 0.447 M

How to determine the mole of ClO₃⁻ in AgClO₃

We'll begin by obtaining the mole of AgClO₃ in the solution. This is illustrated as follow:

  • Volume = 821 mL = 821 / 1000 = 0.821 L
  • Molarity of AgClO₃ = 0.349 M
  • Mole of AgClO₃ =?

Molarity = mole / volume

Cross multiply

Mole = molarity × volume

Mole of AgClO₃ = 0.349 × 0.821

Mole of AgClO₃ = 0.287 mole

Finally, we shall determine the mole of ClO₃⁻ in AgClO₃ solution as folllow:

AgClO₃(aq) <=> Ag⁺(aq) + ClO₃⁻(aq)

From the balanced equation above,

1 mole of AgClO₃ contains 1 mole of ClO₃⁻

Therefore,

0.287 mole of AgClO₃ will also contain 0.287 mole of ClO₃⁻

How to determine the mole of ClO₃⁻ in Mn(ClO₃)₂

We'll begin by obtaining the mole of Mn(ClO₃)₂ in the solution. This is illustrated as follow:

  • Volume = 913 mL = 913 / 1000 = 0.913 L
  • Molarity of Mn(ClO₃)₂ = 0.267 M
  • Mole of Mn(ClO₃)₂ =?

Molarity = mole / volume

Cross multiply

Mole = molarity × volume

Mole of Mn(ClO₃)₂ = 0.267 × 0.913

Mole of Mn(ClO₃)₂ = 0.244 mole

Finally, we shall determine the mole of ClO₃⁻ in Mn(ClO₃)₂ solution as folllow:

Mn(ClO₃)₂(aq) <=> Mn²⁺(aq) + 2ClO₃⁻(aq)

From the balanced equation above,

1 mole of Mn(ClO₃)₂ contains 2 mole of ClO₃⁻

Therefore,

0.244 mole of Mn(ClO₃)₂ will contain = 0.244 × 2 = 0.488 mole of ClO₃⁻

How to determine the molarity of ClO₃⁻ in the resulting solution

  • Mole of ClO₃⁻ in AgClO₃ = 0.287 mole
  • Mole of ClO₃⁻ in Mn(ClO₃)₂ = 0.488 mole
  • Total mole of ClO₃⁻ = 0.287 + 0.488 = 0.775 mole
  • Total volume = 0.821 + 0.913 = 1.734 L
  • Molarity of ClO₃⁻ =?

Molarity = mole / volume

Molarity of ClO₃⁻ = 0.775 / 1.734

Molarity of ClO₃⁻ = 0.447 M

Learn more about molarity:

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