calculate the ph of 0.05 M sulphuric acid when it is fully dissociated.

Dear Aritra

H₂SO₄ -> 2H⁺ + SO4^-2

0.05M will give 0.1 M of H⁺

pH=-log[H⁺]=-log0.1=1

All the best.

AKASH GOYAL

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H₂SO₄ = 2H⁺ + SO₄^2-

 0.05M H₂SO₄ gives 2(0.05)M H⁺

 [H⁺] = 0.1

-log[H⁺] = PH =1