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Question: What is the acidity constant for the following reaction given that the hydronium ion concentration o...

What is the acidity constant for the following reaction given that the hydronium ion concentration of a 0.040.04 M solution of Ni2+N{i^{2 + }} solution of Nickel (II) perchlorate is4.5×10−64.5 \times {10^{ - 6}}?
Ni2+(aq)+2H2O(I)→Ni(OH)+(aq)+H3O+(aq)N{i^{2 + }}(aq) + 2{H_2}O(I) \to Ni{(OH)^ + }(aq) + {H_3}{O^ + }(aq)
a) 2×10−122 \times {10^{ - 12}}
b) 4×10−64 \times {10^{ - 6}}
c) 5×10−125 \times {10^{ - 12}}
d) 5×10−105 \times{10^{ - 10}}

Explanation

Solution

Here Ni2+N{i^{2 + }} in aqueous form, H2O{H_2}O is solvent and H3O+{H_3}{O^ + } is in aqueous form. So we will write the equilibrium equation then we will write the by taking xx as dissociation amount and we will solve for xx.

Complete Step by step answer:
Ni2+(aq)+2H2O(I)→Ni(OH)+(aq)+H3O+(aq)N{i^{2 + }}(aq) + 2{H_2}O(I) \to Ni{(OH)^ + }(aq) + {H_3}{O^ + }(aq)
We are given that the concentration of NiNi is 0.040.04 M respectively.
Let at equilibrium xx amount of reactant has been converted to product so xx amount has been formed in the product side respectively.
So we will write the same equation for equilibrium
                                      Ni2+(aq)+2H2O(I)→Ni(OH)+(aq)+H3O+(aq)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; N{i^{2 + }}(aq) + 2{H_2}O(I) \to Ni{(OH)^ + }(aq) + {H_3}{O^ + }(aq)
At equilibrium (0.04−x)\left( {0.04 - x} \right)                                             \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; xx +                           \;\;\;\;\;\;\;\;\;\;\;\;\; xx
So value of acidity constant Ka{K_a} = concentration  of  productconcentration  of  reactant{\dfrac{concentration \; of \;product}{concentration \; of \;reactant}} = x20.04−x\dfrac{x^2}{0.04 - x}
Here we are given that x=4.5×10−6x = 4.5 \times {10^{ - 6}}
Here the value is <<0.04< < 0.04
Hence we will neglect this
So
Ka=x20.04{K_a} = \dfrac{{{x^2}}}{{0.04}}
⇒Ka=(4.5×10−6)20.04\Rightarrow {K_a} = \dfrac{{{{(4.5 \times {{10}^{ - 6}})}^2}}}{{0.04}}
⇒Ka=20.25×10−120.04\Rightarrow {K_a} = \dfrac{{20.25 \times {{10}^{ - 12}}}}{{0.04}}
⇒Ka=20.25×10−104\Rightarrow {K_a} = \dfrac{{20.25 \times {{10}^{ - 10}}}}{4}
⇒Ka≈5×10−10\Rightarrow {K_a} \approx 5 \times {10^{ - 10}}

Hence the answer is option ‘d’.

Note: While calculating the acidity constant we will keep in mind that the value of xx is usually low so we can ignore this value. So that the calculation will get simple and we will get the answer easily. Also use all the given information in the question wisely in the formulas.