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Question: What are the concentration of \( {H^ + } \) , \( HS{O_4}^ - \) , \( S{O_4}^{2 - } \) and \( {H_2}S{O...

What are the concentration of H+{H^ + } , HSO4−HS{O_4}^ - , SO42−S{O_4}^{2 - } and H2SO4{H_2}S{O_4} in a 0.200.20 M solution of sulphuric acid? Given H2SO4→H++HSO4−{H_2}S{O_4} \to {H^ + } + HS{O_4}^ - ; strong and, HSO4−⇄SO42−; K2=10−2MHS{O_4}^ - \rightleftarrows S{O_4}^{2 - };\,{K_2} = {10^{ - 2}}M

Explanation

Solution

Hint : Equilibrium reactions are those types of reactions, in which all the products formed will react among themselves and lead to the formation of the reactants. For equilibrium to form, the electrolytes should be a week. Weak electrolytes after the dissociation will react among themselves and form the reactants.

Complete Step By Step Answer:
Electrolytes are of two types, strong electrolytes and weak electrolytes. A strong electrolyte will dissociate completely and the products so formed will never react to form the reactants. Like H2SO4→H++HSO4−{H_2}S{O_4} \to {H^ + } + HS{O_4}^ - , whereas a weak electrolyte will never dissociate completely and the products so formed will react with each other to form the reactants. Like HSO4−⇄SO42−; K2=10H−2MHS{O_4}^ - \rightleftarrows S{O_4}^{2 - };\,{K_2} = 10{H^{ - 2}}M .
The given equation is as follows; H2SO4→H++HSO4−{H_2}S{O_4} \to {H^ + } + HS{O_4}^ - and then one of the product HSO4−HS{O_4}^ - further dissociates, HSO4−⇄H++SO42−HS{O_4}^ - \rightleftarrows {H^ + } + S{O_4}^{2 - } .
By the definition of equilibrium constant, we can say that Ka=ProductReactant{K_a} = \dfrac{Product}{Reactant} so now we can write it as Ka=[H+][SO42−][HSO42−]{K_a} = \dfrac{{[{H^ + }][S{O_4}^{2 - }]}}{{[HS{O_4}^{2 - }]}}
But the concentration of the product is not known. Let us take the concentration of the product to be xx and given net concentration is 0.200.20 M thus the concentration remaining for the reactant formed after the dissociation of H2SO4{H_2}S{O_4} is 0.2−x0.2 - x . Now substituting the values to the formula of the equilibrium constant we get, Ka=x×x0.2−x{K_a} = \dfrac{{x \times x}}{{0.2 - x}} ,
On solving the above we get Ka=x20.2−x{K_a} = \dfrac{{{x^2}}}{{0.2 - x}} ,
The given value of Ka{K_a} is 10−2{10^{ - 2}} . now substituting the value of the equilibrium constant to the above equation, we get 10−2=x20.2−x{10^{ - 2}} = \dfrac{{{x^2}}}{{0.2 - x}}
On rearrangement, x2=−10−2x+10−2×0.2{x^2} = - {10^{ - 2}}x + {10^{ - 2}} \times 0.2 or x2+10−2x−10−2×0.2=0{x^2} + {10^{ - 2}}x - {10^{ - 2}} \times 0.2 = 0
On solving the quadratic equation, we get
x=−10−2+(10−2)2+4.1×10−2×0.22x = \dfrac{{ - {{10}^{ - 2}} + \sqrt {{{({{10}^{ - 2}})}^2} + 4.1 \times {{10}^{ - 2}} \times 0.2} }}{2}
⇒x=0.04\Rightarrow x = 0.04
Thus the concentration of [H+]=[SO42−]=0.04M[{H^ + }] = [S{O_4}^{2 - }] = 0.04M
And of [HSO42−]=0.20−0.04=0.16M[HS{O_4}^{2 - }] = 0.20 - 0.04 = 0.16M

Note :
It is to remember that weak electrolytes will always form an equilibrium equation. Whereas strong electrolyte completely dissociates to form the products. While solving any quadratic or cubic or any mathematical equation of any degree try to use their formulas to reduce the calculations and complications.