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8g of potassium chromate is dissolved in acid to make 100cm3 solution. A dynamic equilibrium occurs,

2CrO42-(aq) + 2H+(aq) Cr2O72-(aq) + H2O(l)

(i) What is the Kc expression for above eqiulibrium 

 (ii)calculate pH at which only 1/5 of original amount of chromate ions remain.

6 years ago


Answers : (1)


(i) kc = [Cr2O72-]1[H2O]0  / [CrO42-]2 [H+1]2 = [Cr2O72-]1 / [CrO42-]2 [H+1]2

There is no [H2O] term in the numerator, as the reaction is carried out in aqueous medium. Active mass of any amount of pure liquid or pure solid is 1. So, [H2O]0 =1

(2)Number of mole of pot. chromate = wt /gram molecular wt

     = 8 /194 = 0.04123

From the balanced equation, the No. of mole of H1+ ions required = 0.04123

Concentration of H+1 = No of mole of H+1 / Volume of solution in litres

Concentration of H+1 = 0.04123 / 0.1 = 0.4123 M

pH = log 10 [1 / 0.4123] = log 10 2.425 = 0.3847

6 years ago

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