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Assuming complete ionisation the pH of 0.005M h2so4? |
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Answer» Explanation:GIVEN:H₂SO₄ + 2H₂O ---->2H₃O⁺+ SO₄²⁻0.05 2X0.05pH=-log[0.05Mx 2]pH=-log[0.1M]pH=1∴pH of 0.05M of H2SO4 is 1 |
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