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# 100 mL of sulphuric acid solution (sp. gr. =1.84) contains 98% by weight of pure acid. Calculate the volume of 0.46 M NaOH solution (in L) required to just neutralize the above acid solution.

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Solution

## Density= mass/volume∴ mass=density×volumegiven density of 98% H2SO4=1.84g/mL∴ 1mL of 98% H2SO4 will weigh⇒1.84×1=1.84g∴100mL will weigh ##1.84\times 100=184g\$∴ concentration is given as 98% by mass, 184g of the solution will contain 0.98×184=180.32g of H2SO4molar mass of H2SO4=98.1g∴180.32g constitutes 1.84 molesThe reaction between NaOH and H2SO4 is as follows:2NaOH+H2SO4⟶Na2SO4+2H2O∴ , to neutralize 1 mole of H2SO4,2 moles of NaOH are needed,So, to neutralize 1.84 moles,1.84×2=3.68 moles of NaOHmolar mass of NaOH=40gSo, 3.68 moles constitute 147g of NaOHWe know that 1L of 1M NaOH contains 40g of NaOHSo, 1L of 0.46M NaOH contains 18.4g of NaOH∴ to get 147g of NaOH, we need to take 147/18.4=7.826L of 0.1M NaOH∴ Volume of 0.46M NaOH solution required to neutralize the above acid solution is 7.826L  Suggest Corrections  0      Similar questions
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