A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸

Answer : The concentration of $Ni^{2+}$ ions at equilibrium is, $4.31\times 10^{-8}$

Explanation : Given,

Moles of $NiCl_2$ = 0.130 mol

Volume of solution = 1 L

$Concentration=\frac{Moles }{Volume}$

Concentration of $NiCl_2$ = Concentration of $Ni^{2+}$ = 0.130 M

Concentration of $NH_3$ = 1.20 M

$K_f=5.5\times 10^8$

The equilibrium reaction will be:

$Ni^{2+}(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}$

Initial conc. 0.130 1.20 0

At eqm. x [1.20-6(0.130)] 0.130

= 0.42

The expression for equilibrium constant is:

$K_f=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}$

Now put all the given values in this expression, we get:

$5.5\times 10^8=\frac{(0.130)}{(x)\times (0.42)^6}$

$x=4.31\times 10^{-8}$

Thus, the concentration of $Ni^{2+}$ ions at equilibrium is, $4.31\times 10^{-8}$

7 0

4 years ago

How many grams are needed to form 33.6 L of carbon dioxide

pychu [463]

It depends of the temperature and the pressure. If temperature and pressure are standard 1 mol forms 22.4 liter. So 33.6 liters form 33.6/22.4 = 1.5 mols.

Now calculate the molar mass of carbon dioxide: 12 g/mol + 2x16 g/mol = 44 g/mol

The mass of the 1.5 mols is 1.5mol * 44 g/mol = 66 g.

7 0

3 years ago

A student decided to research primate psychology for their science project. They measured how long it took gorillas to adapt to

Gwar [14]

Answer: the answer is D im pretty sure

Explanation:

3 0

3 years ago

Concentrated hydrochloric acid is 36% (Weight/Volume) hydrochloric acid and has a density of 1.18 g.cm -3.

dezoksy [38]

The concentration refers to the amount of substance that is contained in solution.

<h3>What is the concentration?</h3>

The concentration refers to the amount of substance that is contained in solution. We can be able to obtain the concentration of the raw acid by the use of the relation;

Co = 10pd/M

M = molar mass of the acid

p = percentage of the acid

d = density of the acid

Co = 10 * 36 * 1.18/36.5

Co = 11.6 M

Using the dilution formula;

C1V1 = C2V2

10 * 11.6 = C2 * 1000

C2 = 10 * 11.6/1000

C2 = 0.116 M

Using again;

C1V1 = C2V2

0.116 * 5 = C2 * 20

C2 = 0.116 * 5 /20

C2 = 0.029 M

Learn more about concentration:brainly.com/question/10725862

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8 0

2 years ago