Question

A cyanide solution with avolume of 12.73 mL was treated with 25.00 mL of Ni2+solution (containing

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!