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`        if 100 ml of 1 N H2SO4 is mixed  with 100 ml of 1 M NaOh solution ,the resulting solution  will be`
one year ago

```							The mixture obtained will be neutral solution containing dissolved sodium sulfate. Its pH should, at least theoretically, be equal to 7.The equivalent weight of sulfuric acid, a dibasic acid, is 49 (molecular weight/2). This means that 1 litre of the 1N acid will contain 49 g of H2SO4, and so the amount of acid present in 100 ml is 4.9 g.The molecular weight (formula weight) of sodium hydroxide is 40. It means the amount of NaOH present in 1 litre of a 1M solution is 40 g, and so the amount of the alkali present in 100 ml of such a solution will be 4 g.The equation for the complete neutralisation between the strong acid and the strong base is2NaOH + H2SO4 = Na2SO4 + 2H2OAs per this equation, 2 moles of NaOH (80 g) needs 1 mole of H2SO4 (98 g) for complete neutralisation. From this it follows that 4 g of NaOH needs only 4.9 g of H2SO4 for complete neutralisation, and these amounts of the acid and the alkali also fully tally with the amounts present in 100 ml each of the given acid and alkali solutions.Therefore, when 100 ml of 1N H2SO4 is mixed with 100 ml of 1M sodium hydroxide, we get a neutral solution having a pH value of 7.
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one year ago
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