Chemistry, 23.04.2020 01:17 gildedav001

A solution is prepared by adding 0.10 mole of Ni(NH3)6Cl2 to 0.50 L of 3.0 M NH3. Calculate [Ni(NH3)62 ] and [Ni2 ] in this solution. Koverall for Ni(NH3)62 is 5.5 x 108. That is, 5.5 x 108

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Answer from: kmassarotti

$[Ni^{2+}]= 2.327*10^{-8} M$

$[Ni(NH_3)_6]^{2+} _{(aq)}]= 0.3 M$

Explanation:

From the question we are told that

The number of moles of Ni(NH_3)_6Cl_2 is $n = 0.10 \ moles$

The volume of NH_3 is = V = 0.50L

The concentration of NH_3 C = 3.0M

The $k_{overall}$ for $[Ni(NH_3) _6^{2+}]$ is $k_{overall} = 5.5*10^{8}$

The number of moles of NH_3 is mathematically evaluated as

$No\ of mole = concentration \ * \ volume$

substituting values

Moles = 0.50* 3.0

$=1.5 \ moles$

The equation for this reaction is

$Ni^{2+}_{(aq)} + 6NH_3_{(aq)}$ ⇄ $[Ni(NH_3)_6]^{2+} _{(aq)}$

From this equation we can mathematically evaluate the limiting reactant as

For $Ni^{2+}$

$0.10moles [Ni ^{2+}] * \frac{6moles NH_3}{1 mole [N_i^{2+}]}$

$= 0.6 moles [Ni^{2+}]$

For NH_3

$1.5 moles [NH_3] * \frac{6moles NH_3}{1 mole [N_i^{2+}]}$

= 9 moles [NH_3]

Hence the Limiting reactant is $Ni^{2+}$

At the initial the number of moles of $[Ni(NH_3)_6]^{2+} _{(aq)}$ produced is mathematically evaluated as

$0.10 [Ni^{2+}] *\frac{1 moles [Ni(NH_3)_6]^{2+}}{1 mol [Ni^{2+}]}$

$= 0.10 \ moles$

The remaining amount of NH_3 is $1.5 - 0.6 = 0.9 \ moles$

Now the reaction is complete

The number of moles of $Ni^{2+}$ = 0 moles

For this kind of reaction after the complex solution(which is at equilibrium) has been formed they disassociate again ( which is also at equilibrium )

Let the number of moles of $Ni^{2+}$ after disassociation be z

The the number of moles of NH_3 after disassociation be 0.9 + 6z

The 6 is from the reaction equation

The the number of moles of $[Ni(NH_3)_6]^{2+} _{(aq)}$ after disassociation be 0.10 -z

Now the equilibrium constant for this reaction is

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

The powers of these concentration is are their moles

substituting value

$5.5*10^{8} = \frac{0.10 - z}{[z] [0.9 + 6z]^6}$

Since the back reaction is very little so we can neglecting subtracting it from the moles of $[Ni(NH_3)_6]^{2+} _{(aq)}$ and adding it to the number of moles of NH_3

Because it won't affect the value of z obtained

$5.5*10^{8} = \frac{0.10}{[z] [0.5]^6}$

$z = \frac{0.10}{5.5*10^8 * (0.9)}$

$= 1.16*10^{-8} \ mol$

So the number of moles of $Ni^{2+}$ at equilibrium is $= 1.16*10^{-8} \ mol$

The concentration of $Ni^{2+}$ is mathematically represented as

$concentration = \frac{mole}{volume}$

$= \frac{1.1636*10^{-8}}{0.50}$

$[Ni^{2+}]= 2.327*10^{-8} M$

The moles of $[Ni(NH_3)_6]^{2+} _{(aq)}$ is = $0.10 - 2.327*10^{-8}$

$= 0.10 \ moles$

The concentration of $[Ni(NH_3)_6]^{2+} _{(aq)}$ is mathematically represented as

$concentration = \frac{mole}{volume}$

$= \frac{0.10}{0.50}$

$[Ni(NH_3)_6]^{2+} _{(aq)}]= 0.3 M$

Answer from: Quest

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

Answer from: Quest

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