Home
Class 11
CHEMISTRY
The pH of a solution of H(2)SO(4) is 1. ...

The pH of a solution of `H_(2)SO_(4)` is 1. Assuming complete ionisation, find the molarity of `H_(2)SO_(4)` solution :

A

`0.1`

B

`0.2`

C

`0.05`

D

`2.0`

Text Solution

Verified by Experts

Promotional Banner

Topper's Solved these Questions

  • IONIC EEQUILIBRIUM

    NARENDRA AWASTHI|Exercise Level- 1|1 Videos
  • IONIC EEQUILIBRIUM

    NARENDRA AWASTHI|Exercise Level- 2|10 Videos
  • GASEOUS STATE

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|1 Videos
  • SOLID STATE

    NARENDRA AWASTHI|Exercise Level 3 - Subjective Problems|1 Videos

Similar Questions

Explore conceptually related problems

Gram molecular mass of H_(2)SO_(4) is

What is the normality of 1.5 M H_2SO_4 .

What is the molartiy of SO_4^(2-) ion in aqueous solution that contain 34.2 ppm of AI_2(SO_4)_3 ? (Assume complete dissociation and density of solution 1g/mL )

A 3.4g sample of H_(2)O_(2) solution containing x%H_(2)O by mass requires xmL of a KMnO_(4) solution for complete oxidation under acidic conditions. The normality of KMnO_(4) solution is :

FeS + H_2SO_4 rarr

Find the molarity and molality of a 14% solution of H_(2) SO_(4) by weight. The density of the solution is 1.10 g/cc and the molecular mass of sulphuric acid is 98

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) If excess water is added into a bottle sample labelled as "112% H_(2)SO_(4) " and is reacted with 5.3 g NaCO_(3) then find the volume of CO_(2) evolved at 1 atm pressure and 300 K temperature after the completion of the reaction :

Oleum is considered as a solution of SO_(3) in H_(2)SO_(4) , which is obtained by passing SO_(3) in solution of H_(2)SO_(4) When 100 g sample of oleum is diluted with desired mass of H_(2)O then the total mass of H_(2)SO_(4) obtained after dilution is known is known as % labelling in oleum. For example, a oleum bottle labelled as ' 019% H_(2)SO_(4) ' means the 109 g total mass of pure H_(2)SO_(4) will be formed when 100 g of oleum is diluted by 9 g of H_(2)O which combines with all the free SO_(3) present in oleum to form H_(2)SO_(4) as SO_(3)+H_(2)O to H_(2)SO_(4) 1 g of oleum sample is diluted with water. The solution required 54 mL of 0.4 N NaOH for complete neutralization. The % free SO_(3) in the sample is :

when an aqueous solution of H_2SO_4 is electrolysed the product at anodes is :