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Calculate the pH of a solution obtained by dissolving 490 mg H2SO4 in water so that the total volume of solution becomes 100 ml . ( please explain the derivation)
First of all calculate no. Of moles of sulfuric acid added by div given mass by molar mass of acid that is 98 Then find concentration simply by formula Then use pH = - log(H+) to get the answer as=log(-0.05) I think it would work
Molecular mass of H2 SO4 Sulfuric acid = 98Num of moles = 0.490 /98 = 0.005Volume of solution = 0.100 LMolarity = 0.005/0.100 = 0.050 MThere are 2 ions of H+ for each molecule of the acid. So[ H⁺ ] = 0.100 MpH = - Log₁₀ [ H⁺ ] = 1
First calculate normality which gives the concentration of Hydrogen ion and then calculate pH using equation
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