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Suppose a 50.0 ml of 0.250 M CoCl2 solution is added to 25.0 mL of 0.350 M NiCl2 solution. Calculate the concentration, in moles per liter, of each of the ions present after mixing. Assume that the volumes are additive.

User Ndtreviv
by
6.9k points

2 Answers

2 votes

Answer:

The answer to your question is below

Cobalt M = 0.17

Chlorine M = 0.0213

Nickel M = 0.12 M

Step-by-step explanation:

Data

50 ml of CoCl2 0.250 M

25 ml of NiCl2 0.350 M

Molarity =
( moles)/(volume)

For Cobalt and Chlorine

moles = Molarity x volume

moles = 0.250 x 0.050

moles = 0.0125

For nickel and chlorine

moles = 0.350 x 0.025

moles = 0.0088

Total volume = 0.050 + 0.025

Cobalt

M
= (0.125)/(0.075) = 0.17

Chlorine

total moles = 0.0125 + 0.0088

total moles = 0.0213

M =
(0.0213)/(0.075) = 0.28 M

Nickel

M =
(0.0088)/(0.075) = 0.12

User Johnny Baloney
by
7.1k points
3 votes

Answer:

Co⁺² = 0.167 M

Ni⁺² = 0.117 M

Cl⁻ = 0.567 M

Step-by-step explanation:

When the solutions are mixed, the Ni and Co will change between them, so the only salts that can be produced are CoCl₂ and NiCl₂, which are the initial ones. The question already said that they are in aqueous solution, so we can conclude that they are soluble, and after the mixing, there'll be no precipitated formed.

The dissolution happens by the equations below:

CoCl₂ → Co⁺² + 2Cl⁻

NiCl₂ → Ni⁺² + 2Cl⁻

By the stoichiometry of the reactions, the initial molarity of the ions are:

CoCl₂: Co⁺² = 0.250 M, Cl⁻ = 2*0.250 = 0.500 M

NiCl₂: Ni⁺² = 0.350 M, Cl⁻ = 2*0.350 = 0.700 M

The number of moles is the molarity multiplied by the volume in L (0.05 L, and 0.025 L):

nCo⁺² = 0.250*0.05 = 0.0125 mol

nNi⁺² = 0.350*0.025 = 0.00875 mol

nCl⁻ = (0.500*0.05) + (0.700*0.025) = 0.0425 mol

After the mixing, the volume will be 75 mL = 0.075 L, so the concentration will be the number of moles divided by the volume:

Co⁺² = 0.0125/0.075 = 0.167 M

Ni⁺² = 0.00875/0.075 = 0.117 M

Cl⁻ = 0.0425/0.075 = 0.567 M

User Kelly Ethridge
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