Question

A cyanide solution with avolume of 12.73 mL was treated with 25.00 mL of Ni2+solution (containing
excess Ni2+) to convert the cyanide totetracyanonickelate(II):
4CN- + Ni2+ ------Ni(CN)42-
The excess Ni2+ was then titrated with 10.15 mL of0.01307 M ethylenediaminetetraacetic acid
(EDTA). One mole of this reagent reacts with 1 mol ofNi2+:
Ni2+ + EDTA4- ----Ni(EDTA)2-
Ni(CN)42- does not react with EDTA. If 39.35mL of the 0.01307 M EDTA was required to react
with 30.10 mL of the original Ni2+ solution, calculatethe molarity of CN- in the 12.73 mL cyanide
sample.

Answer 1

Answer:

The molarity of this solution is 0,09254M

Explanation:

The concentration of the Ni²⁺ solution is:

Ni²⁺ + EDTA⁴⁻ → Ni(EDTA)²⁻

0,03935L × 0,01307M = 5,143x10⁻⁴ moles Ni²⁺ ÷ 0,03010L =0,01709M Ni²⁺

25,00 mL of this solution contain:

0,01709M × 0,02500L = 4,2716x10⁻⁴ moles of Ni²⁺

The moles of Ni²⁺ that are in excess and react with EDTA⁴⁻ are:

0,01015L × 0,01307M = 1,3266x10⁻⁴ moles of Ni²⁺

Thus, moles of Ni²⁺ that react with CN⁻ are:

4,2716x10⁻⁴ - 1,3266x10⁻⁴ = 2,9450x10⁻⁴ moles of Ni²⁺

For the reaction:

4CN⁻ + Ni²⁺ → Ni(CN)₄²⁻

Four moles of CN⁻ react with 1 mole of Ni²⁺:

2,9450x10⁻⁴ moles of Ni²⁺ × $\frac{4 mol CN^-}{1 molNi^{2+}}$ = 1,178x10⁻³ moles of CN⁻

As the volume of cyanide solution is 12,73mL. The molarity of this solution is:

1,178x10⁻³ moles of CN⁻ ÷ 0,01273L = 0,09254M

I hope it helps!

Answer 2

Based on the data provided, the molarity of C_N⁻ in the cyanide solution is 0.09254 M

What is the molarity of a solution?

The molarity of a solution is the amount in moles of a solute present in a given volume of solution.

• Molarity = moles/volume

The concentration of the Ni²⁺ ion solution is first determined from the equation of the reaction.

Equation of the reaction is given below:

• Ni²⁺ + EDTA⁴⁻ → Ni(EDTA)²⁻

From the equation of the reaction, 1 mole of Ni²⁺ reacts with 1 mole of EDTA

Moles of EDTA = Moles of Ni²⁺

Moles of EDTA = 0.03935L × 0.01307 M = 5.143x10⁻⁴ moles Ni²⁺

Molarity of Ni²⁺ = 5.143x10⁻⁴ moles Ni²⁺/0.03010L

Molarity of Ni²⁺ = 0.01709 M Ni²⁺ solution

25.00 mL of this solution contains:

0.01709M × 0.02500L = 4.2716 × 10⁻⁴ moles of Ni²⁺

The moles of Ni²⁺ that are in excess which will react with EDTA⁴⁻ is:

0.01015L × 0.01307M = 1.3266x10⁻⁴ moles of Ni²⁺

Then, moles of Ni²⁺ that react with C_N⁻ will be:

4.2716x10⁻⁴ - 1.3266x10⁻⁴ = 2.9450 × 10⁻⁴ moles of Ni²⁺

The equation of the reaction of C_N⁻ and Ni² is given below:

• 4C_N⁻ + Ni²⁺ → Ni(C_N)₄²⁻

From the equation of the reaction, 4 moles of C_N⁻ react with 1 mole of Ni²⁺

• 2.9450 × 10⁻⁴ moles of Ni²⁺ will react with 2.9450 × 10⁻⁴ moles × 4 moles of C_N⁻ = 1.178x10⁻³ moles of C_N⁻

• Molarity of cyanide solution = moles/volume

Volume of cyanide solution is 12.73mL

molarity of cyanide solution = 1.178x10⁻³/0.01273L = 0.09254 M

Therefore, the molarity of C_N⁻ in the cyanide solution is 0.09254 M

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