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.

(i) k_c = [Cr₂O₇^2-]¹[H₂O]⁰  / [CrO₄^2-]² [H⁺¹]² = [Cr₂O₇^2-]¹ / [CrO₄^2-]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