Home
Class 12
CHEMISTRY
Dilute H(2)SO(4) can be used in group o...

Dilute `H_(2)SO_(4)` can be used in group of dil HCI. True/False

Text Solution

AI Generated Solution

The correct Answer is:
To determine whether dilute \( H_2SO_4 \) can be used in the group of dilute \( HCl \), we need to analyze the behavior of these acids in the context of qualitative inorganic salt analysis. ### Step-by-Step Solution: 1. **Understanding the Role of Acids in Salt Analysis:** - In qualitative analysis, acids are used to dissolve salts and to provide a medium for the precipitation of certain ions. - Dilute hydrochloric acid (\( HCl \)) is commonly used because it provides \( H^+ \) ions and chloride ions. 2. **Common Ion Effect:** - When \( HCl \) is present, it provides \( H^+ \) ions, which can affect the ionization of other acids, such as hydrogen sulfide (\( H_2S \)). - The presence of common \( H^+ \) ions from \( HCl \) decreases the ionization of \( H_2S \), leading to fewer sulfide ions available for precipitation. 3. **Properties of Dilute \( H_2SO_4 \):** - Dilute sulfuric acid (\( H_2SO_4 \)) is also a strong acid and a good source of \( H^+ \) ions. - It can also precipitate certain metal ions (like lead, zinc, barium, calcium, cerium) as their sulfates. 4. **Comparison of \( HCl \) and \( H_2SO_4 \):** - While both acids provide \( H^+ \) ions, the presence of sulfate ions from \( H_2SO_4 \) can lead to the precipitation of different metal ions compared to \( HCl \). - However, the key point is that \( H_2SO_4 \) does not interfere with the analysis of Group II sulfides in the same way that \( HCl \) does. 5. **Conclusion:** - Since dilute \( H_2SO_4 \) can be used without significantly affecting the ionization of sulfides and can also precipitate certain metal ions, the statement that dilute \( H_2SO_4 \) can be used in the group of dilute \( HCl \) is **True**. ### Final Answer: **True**

To determine whether dilute \( H_2SO_4 \) can be used in the group of dilute \( HCl \), we need to analyze the behavior of these acids in the context of qualitative inorganic salt analysis. ### Step-by-Step Solution: 1. **Understanding the Role of Acids in Salt Analysis:** - In qualitative analysis, acids are used to dissolve salts and to provide a medium for the precipitation of certain ions. - Dilute hydrochloric acid (\( HCl \)) is commonly used because it provides \( H^+ \) ions and chloride ions. ...
Promotional Banner

Topper's Solved these Questions

  • QUALITATIVE INORGANIC SALT ANALYSIS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises Archives (Linked Comprehension)|2 Videos
  • QUALITATIVE INORGANIC SALT ANALYSIS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises Archives (Multiple Correct)|4 Videos
  • QUALITATIVE INORGANIC SALT ANALYSIS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Fill In The Blanks)|42 Videos
  • P-BLOCK GROUP 18 ELEMENTS - THE INERT GASES

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 5.1 (Objective)|14 Videos
  • REDUCTION AND OXIDATION REACTION OF ORGANIC COMPOUNDS

    CENGAGE CHEMISTRY ENGLISH|Exercise SUBJECTIVE TYPE|4 Videos

Similar Questions

Explore conceptually related problems

State True or false : concentrated H_(2)SO_(4) can not be used for drying H_(2) ?

concentrated H_2 SO_4 can be used to dry which gas ?

During the electrolysis of dilute H_(2)SO_(4) oxygen is liberated at anode . Give reason .

The dilute aqueous solution of H_(2)O_(2) turns blue litmus red. True/False

Radium is a radioactive substance. It dissolves in dilute H_(2)SO_(4) and forms a compound radium sulphate. The compound is

Volumetric titrations involving KMnO_(4) , are carried out only in presence of dilute H_(2)SO_(4) but not in the presence of HCI or HNO_(3) This is because oxygen produced from KMnO_(4) + dil: H_(2)SO_(4) , is used only for oxidizing the reducing agent. Moreover, H2SO4 does not give any oxygen of its own to oxidize the reducing agent. In case HCI is used, the oxygen produced from KMnO_(4) + HCI is partly used up to oxidize HCI and in case HNO_(3) is used, it itself acts as oxidizing agent and partly oxidizes the reducing agent. KMnO_(4) , in various mediums gives following products Mn^(7+)overset(H^(+))toMn^(2+) Mn^(7+)overset(H_(2)O)toMn^(4+) Mn^(7+)underset(OH^(-))toMn^(6+) Q MnO_(4)^(2-) (1 mole) in neutral medium disproportionates to,

Volumetric titrations involving KMnO_(4) , are carried out only in presence of dilute H_(2)SO_(4) but not in the presence of HCI or HNO_(3) This is because oxygen produced from KMnO_(4) + dil: H_(2)SO_(4) , is used only for oxidizing the reducing agent. Moreover, H2SO4 does not give any oxygen of its own to oxidize the reducing agent. In case HCI is used, the oxygen produced from KMnO_(4) + HCI is partly used up to oxidize HCI and in case HNO_(3) is used, it itself acts as oxidizing agent and partly oxidizes the reducing agent. KMnO_(4) , in various mediums gives following products Mn^(7+)overset(H^(+))toMn^(2+) Mn^(7+)overset(H_(2)O)toMn^(4+) Mn^(7+)underset(OH^(-))toMn^(6+) QThe number of moles of KMnO_(4) that will be needed to react with one mole of sulphite in an acidic solution is

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 109% 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) What is the % of free SO_(3) in an oleum that is labelled as 104.5% H_(2)SO_(4) ?

Volumetric titrations involving KMnO_(4) , are carried out only in presence of dilute H_(2)SO_(4) but not in the presence of HCI or HNO_(3) This is because oxygen produced from KMnO_(4) + dil: H_(2)SO_(4) , is used only for oxidizing the reducing agent. Moreover, H2SO4 does not give any oxygen of its own to oxidize the reducing agent. In case HCI is used, the oxygen produced from KMnO_(4) + HCI is partly used up to oxidize HCI and in case HNO_(3) is used, it itself acts as oxidizing agent and partly oxidizes the reducing agent. KMnO_(4) , in various mediums gives following products Mn^(7+)overset(H^(+))toMn^(2+) Mn^(7+)overset(H_(2)O)toMn^(4+) Mn^(7+)underset(OH^(-))toMn^(6+) QThe equivalent weight of KMnO_(4) , in the given reaction is KMnO_(4)toK_(2)MnO_(4)+ MnO_(2) + O_(2)

Volumetric titrations involving KMnO_(4) , are carried out only in presence of dilute H_(2)SO_(4) but not in the presence of HCI or HNO_(3) This is because oxygen produced from KMnO_(4) + dil: H_(2)SO_(4) , is used only for oxidizing the reducing agent. Moreover, H2SO4 does not give any oxygen of its own to oxidize the reducing agent. In case HCI is used, the oxygen produced from KMnO_(4) + HCI is partly used up to oxidize HCI and in case HNO_(3) is used, it itself acts as oxidizing agent and partly oxidizes the reducing agent. KMnO_(4) , in various mediums gives following products Mn^(7+)overset(H^(+))toMn^(2+) Mn^(7+)overset(H_(2)O)toMn^(4+) Mn^(7+)underset(OH^(-))toMn^(6+) QNumber of electrons transferred in each case when KMnO_(4) acts as an oxidizing agent to give MnO_(2) , Mn^(++) Mn(OH)_(3) , and MnO_(4)^(2-) are respectively

CENGAGE CHEMISTRY ENGLISH-QUALITATIVE INORGANIC SALT ANALYSIS-Exercises (True And False )
  1. Turnbull's blue and prussian's blue respectively are KFe^(II)[Fe^(...

    Text Solution

    |

  2. K(sp) of Mg(OH)(2) is 1xx10^(-12), 0.01M MgCI(2) will be precipitating...

    Text Solution

    |

  3. There is ppt. of solute AB if its product is greater than K(sp) val...

    Text Solution

    |

  4. Chlorine gas is passed into a solution containing KF, KI and KBr and C...

    Text Solution

    |

  5. When H(2)S gas is passed into aq ZnCl(2) solution white ppt of ZnS is...

    Text Solution

    |

  6. Dilute H(2)SO(4) can be used in group of dil HCI. True/False

    Text Solution

    |

  7. NH(4)CI can be replaced by (NH(4))(2)SO(4) in group III.

    Text Solution

    |

  8. Alkaline solution of NH(4)Cl gives ppt with K(2)HgI(4)

    Text Solution

    |

  9. When KNO(2) and CH(2)COOH is added as CoCl(2) solution, yellow ppt of ...

    Text Solution

    |

  10. K(4)[Fe(CN)(6)] is used to test Cu^(2+),Fe^(2+),Zn^(2+),Cd^(2+)

    Text Solution

    |

  11. Hg(2)Cl(2) is black ened by NH(3) due to formation of iodide of mi...

    Text Solution

    |

  12. White ppt of PbCl(2) is soluble in aq NH(3). (T/F)

    Text Solution

    |

  13. If acidified solution of K(2)Cr(2)O(7) turm green on addition of...

    Text Solution

    |

  14. In group II, Formqation of whichsh tarbidity on dilation with H(2)O ...

    Text Solution

    |

  15. NaOH can be used to seprate Al(OH)(3) and Zn(OH)(2) . (T/F)

    Text Solution

    |

  16. NH(4)SCN can be used to make distanction between Cu^(2+) and Co^(2+)...

    Text Solution

    |

  17. Yellow ammonium sulphide (YAS) can be used to seprate SnS and As(2)S...

    Text Solution

    |

  18. NaOH can be used to seprate Al(OH)(3) and Zn(OH)(2) . (T/F)

    Text Solution

    |

  19. AlCl(2) is soluble is axcess of NaOH forming sodium metaaluminate ...

    Text Solution

    |

  20. BaBr(2) gives yellow ppt with AgNO(3) as well as with K(2)CrO(4). (T...

    Text Solution

    |