Question

What volume of 0.1 M NaOH solution is required to neutralize 100 ml of
concentrated aqueous solution of sulphuric acid which contains 98% sulphuric acid by mass.
The density of concentrated sulphuric acid is 1.84g/cc. NaOH reacts with sulphuric
acid according to the following reaction.
2NaOH+H2SO4Na2SO4+2H2O

Answer 1

Explanation:

CHEMISTRY

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.

December 27, 2019avatar

Viraj Rohilla

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ANSWER

Density= mass/volume

∴ mass=density×volume

given density of 98% H

2

SO

4

=1.84g/mL

∴ 1mL of 98% H

2

SO

4

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 H2so4.

molar mass of H2so4 =98.1g

∴180.32g constitutes 1.84 moles

The reaction between NaOH and H2so4

is as follows:

2NaOH+H2so4

∴ , 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 NaOH

molar mass of NaOH=40g

So, 3.68 moles constitute 147g of NaOH

We know that 1L of 1M NaOH contains 40g of NaOH

So, 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