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

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Answer from: Quest

explanation: no but i do not usually comment

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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)...

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