Home
Class 11
CHEMISTRY
5 mL of 8 N HNO(3), 4.8 mL of 5 N HCl, a...

`5 mL` of `8 N HNO_(3), 4.8 mL` of `5 N HCl`, and a certain volume of `17 m H_(2) SO_(4)` are mixed together and made upto `2 L`. `30 mL` of the acid mixture exactly neutralises `42.9 mL` of `Na_(2 CO_(3)` solution containing `0.1 g` of `Na_(2) CO_(3). 10 H_(2) O` in `10 mL` of water. Calculate:
The amount (in g) of the sulphate ions in the solution.

Text Solution

Verified by Experts

mEq of acid mixture
`= mEq of HNO_(3) + mEq of HCl + mEq of H_(2) SO_(4)`
Let `N` be the normality of the acid mixture and `V mL` be the volume of `H_(2) SO_(4)` added.
`N xx 200 = 8 xx 5 xx 4.8 + 2 (17) xx V`
Now find `N` of carbonate as follows:
`implies N = ("Strength")/(Ew)`
Strength `= 0.1//10 mL -= 10 g L^(-1)`
`Ew = Mw//2 = 286//2 = 143`
(`Mw = 106 + 180`, adding the mass of `10 H_(2) O`)
`N = (10)/(143)`
Now mEq of acid mixture = mEq of `Na_(2)CO_(3)` solution `N xx 30 = 42.9 xx (10)/(143)`
`implies N = 0.1 =` Normality of acid mixture
Substituting in equation (i) we get:
`0.1 xx 2000 = (40 + 24 + 34) V_(mL)`
For grams of sulphate ions:
mEq of `H_(2) SO_(4) = 2 xx 17 xx V = 136` `(VmL = 4)`
Now, mEq of `SO_(4)^(2-) = mEq of H_(2) SO_(4)`
`("Weight")/(E_(SO_(4)^(2-))) xx 1000 = 136`
`("Weight")/(96//2) xx 1000 = 136`
Weight = Grams of `SO_(4)^(2-)` ions `= 6.53 g`
Promotional Banner

Topper's Solved these Questions

  • SOME BASIC CONCEPTS AND MOLE CONCEPT

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 1.1 Objective Questions|6 Videos

  • SOME BASIC CONCEPTS AND MOLE CONCEPT

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 1.1 Objective Questions (Single Correct)|3 Videos

  • S-BLOCK GROUP 2 - ALKALINE EARTH METALS

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 5.1 Objective|2 Videos

  • STATES OF MATTER

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Ture False)|25 Videos

Similar Questions

Explore conceptually related problems

10 " mL of " 2 M HCl and 20 " mL of " 1 M HNO_3 and V volume of 5M H_2SO_4 are mixed together and the solution was made upto 5 L. 10 " mL of " this acid solution exactly neutralises 28.6 " mL of " Na_2CO_3 solution containing 1 g of Na_2CO_4.10H_2O in 100 " mL of " water. Calculate the amount of SO_4^(2-) ions in grams present in solution.

50 mL of 1 M HCl, 100 mL of 0.5 M HNO_(3) , and x mL of 5 M H_(2)SO_(4) are mixed together and the total volume is made upto 1.0 L with water. 100 mL of this solution exactly neutralises 10 mL of M//3 Al_(2) (CO_(3))_(3) . Calculate the value of x .

25mL of 2N HCI, 50 mL of 4N HNO_3 and x mL of 2MH_2 SO_4 are mixed together and the total volume is made up to 1L after dilution. 50mL of this acid mixture completely reacted with 25mL of a 1NNa_2 CO_3 solution. The value of x is :

25mL of 2N HCl, 50 mL "of" 4N HNO_(3) and x mL 2 M H_(2)SO_(4) are mixed together and the total volume is made up to 1L after dilution. 50mL if this acid ixture completely reacteed with 25mL of a 1 N Na_(2)CO_(3) solution. The value of x is:

4.0 g of NaOH and 4.9 g of H_(2)SO_(4) are dissolved in water and volume is made upto 250 mL. The pH of this solution is:

5 mL of N HCI, 20 mL of N/2 H_2SO_4 and 30 mL of N/3 HNO_3 are mixed together and the volume made to 1 litre. The normality of the resulting solution is:

5 mL of N-HCl , 20 mL of N//2 H_(2)SO_(4) and 30 mL of N//3 HNO_(3) are mixed together and the volume is made to 1 L . The normality of the resulting solution is

20 mL N/2 HCI, 60 mL N/10 H_2SO_4 and 150 mL N/5 HNO_3 are mixed. Calculate the normality of the mixture of acids in solution.

20ml of "0.5 M KMnO"_(4) in acidic media exactly neutralises 30ml of a H_(2)O_(2) solution. The volume strength of H_(2)O_(2) solution will be?

500 mL fo 2 M HCl , 100 mL of 2 M H_(2) SO_(4) , and one gram equivalent of a monoacidic alkali are mixed together. 30mL of this solution requried 20 mL of 143 g Na_(2) CO_(3). xH_(2)O in one litre solution. Calculate the water of crystallisation of Na_(2) CO_(3). xH_(2) O