Acyanide 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+ (cn)42-
the excess ni2+ was then titrated with 10.15 ml of0.01307 m acid
(edta). one mole of this reagent reacts with 1 mol ofni2+:
ni2+ + edta4- (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.

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⁺ × = 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!

the answer for sodium is 0.186

the answer for rubidium is 0.251