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Question: A monoprotic acid in \(1.00\)M solution is \(0.001\) % ionised. The dissociation constant of acid is...

A monoprotic acid in 1.001.00M solution is 0.0010.001 % ionised. The dissociation constant of acid is:
A. 1×1031 \times \,{10^{ - 3}}
B. 1×1061 \times \,{10^{ - 6}}
C. 1×1081 \times \,{10^{ - 8}}
D. 1×10101 \times \,{10^{ - 10}}

Explanation

Solution

The dissociation constant is determined as the product of the square of the degree of dissociation and concentration. We will use the dissociation constant formula to determine the dissociation constant. It is given that the acid is monoprotic so the van’t Hoff factor will remain one. We can use the simple dissociation constant formula.

Formula used: Ka=Cα21α{{{K}}_{{a}}}{{ = }}\,\dfrac{{{{C}}{{{\alpha }}^{{2}}}}}{{{{1}} - {{\alpha }}}}

Complete step-by-step answer:
The formula which relates the degree of dissociation with dissociation constant is as follows:
Ka=Cα21α{{{K}}_{{a}}}{{ = }}\,\dfrac{{{{C}}{{{\alpha }}^{{2}}}}}{{{{1}} - {{\alpha }}}}
Where,
Ka{{{K}}_{{a}}}is the acid dissociation constant.
C is the concentration.
α{{\alpha }}is the degree of dissociation
1.001.00M solution is 0.0010.001 % ionised

On substituting 1.001.00M for the concentration and 0.0010.001 for degree of dissociation.
Ka=1.00×(0.001)210.001\Rightarrow {{{K}}_{{a}}}{{ = }}\,\dfrac{{{{1}}{{.00}}\, \times {{{{(0}}{{.001)}}}^{{2}}}}}{{{{1}} - 0.001}}
Ka=1×1060.99\Rightarrow {{{K}}_{{a}}}{{ = }}\,\dfrac{{{{1}} \times {{1}}{{{0}}^{ - 6}}}}{{0.99}}
Ka=1×106\Rightarrow {{{K}}_{{a}}}{{ = }}\,{{1}} \times {{1}}{{{0}}^{ - 6}}

Mostly weak electrolyte dissociates very less so, for weak electrolyte the degree of dissociation is very-very less than one,α<<1\alpha < < 1 so, the value of 1α1 - \alpha can be taken to equal to11so, the formula of the dissociation constant can be reduced as, Ka=Cα2{K_a} = \,C{\alpha ^2}.
Ka=1.00×(0.001)2\Rightarrow {{{K}}_{{a}}}{{ = }}\,{{1}}{{.00}}\, \times {{{(0}}{{.001)}}^{{2}}}
Ka=1×106\Rightarrow {{{K}}_{{a}}}{{ = }}\,{{1}} \times {{1}}{{{0}}^{ - 6}}
So, the dissociation constant of acid is 1×106{{1}} \times {{1}}{{{0}}^{ - 6}}.

Therefore, option (B) 1×1061 \times \,{10^{ - 6}}, is correct.

Note: The degree of dissociation tells the dissociated amount of the weak electrolyte. The strong electrolyte dissociates completely, so it is not calculated for the strong electrolyte. A similar formula is used for the determination of the degree of dissociation for a weak base only theKa{{{K}}_{{a}}} is replaced with Kb{{{K}}_{{b}}}where, Kb{{{K}}_{{b}}} is the base dissociation constant. The degree of dissociation can also be calculated by using equivalent conductance at a concentration and equivalent conductance at infinite dilution. The formula to calculate the degree of dissociation is α=λmλm\alpha = \,\dfrac{{{\lambda _m}}}{{\lambda _m^ \circ }}. Where, λm{\lambda _m} is the equivalent conductance at a concentration and λm\lambda _m^ \circ is the equivalent conductance at infinite dilution.