What is pH of 0.05 M sulphuric acid?
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Given:
H₂SO₄ + 2H₂O ---->2H₃O⁺+ SO₄²⁻
0.05 2X0.05
pH=-log[0.05Mx 2]
pH=-log[0.1M]
pH=1
∴pH of 0.05M of H2SO4 is 1
H₂SO₄ + 2H₂O ---->2H₃O⁺+ SO₄²⁻
0.05 2X0.05
pH=-log[0.05Mx 2]
pH=-log[0.1M]
pH=1
∴pH of 0.05M of H2SO4 is 1
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