Home
Class 12
CHEMISTRY
In an experiment 50ml of 0.1(M) solution...

In an experiment `50ml` of `0.1(M)` solution of a salt is reacted with `25ml` of `0.1(M)` solution of sodium sulphite. The half equation for the oxidation of sulphite ion is `SO_(3)^(2-)(aq)+H_(2)O rarr SO_(4)^(2-)(aq)+2H^(+)(aq)+2e^(-)` If the oxidation number of metal in the salt was `3`, what would be the new oxidation number of metal?

Text Solution

Verified by Experts

The correct Answer is:
2
meq of `Na_(2)SO_(3)=` meq of salt
`25xx0.1xx2=50xx0.1xxn` factor.
Promotional Banner

Topper's Solved these Questions

  • TEST PAPERS

    RESONANCE|Exercise PART - III CHEMISTRY SEC - 1|12 Videos

  • TEST PAPERS

    RESONANCE|Exercise PART - III CHEMISTRY SEC - 2|18 Videos

  • SURFACE CHEMISTRY

    RESONANCE|Exercise Section - 5|1 Videos

  • TEST SERIES

    RESONANCE|Exercise CHEMISTRY|50 Videos

Similar Questions

Explore conceptually related problems

50 mL of 0.1 M solution of a salt reacted with 25 mL of 0.1 M solution of sodium sulphite. The half reaction for the oxidation of sulphite ion is: SO_(3)^(2-)(aq)+H_(2)O(l) rarr (aq)+2H^(+)(aq)+2e^(-) If the oxidation number of metal in the salt was 3 , what would be the new oxidation number of metal:

50 mL of 0.1M solution of a salt reacted with 25 mL of 0.1M solution of sodium sulphite. The half reaction for the oxidation of sulphate ion is: SO_(3(aq.))^(2-)+H_(2)O_((l))rarrSO_(4(aq.))^(2-)+2H_(+(aq.))+2e If the oxidation number of metal in the salt was 3 , what would be the new oxidation number of metal?

20 ml of 0.1 M solution of metal ion reacted with 20 ml to 0.1 M SO_(2) solution . SO_(2) reacted according to the equation SO_(2) + 2H_(2)O rarr SO_(4)^(2-) + 4H^(+) + 2e^(-) . If the oxidation no. of metal ion was + 3 , the new oxidation number of the metal would be :

In an experiment, 50 mL of 0.1 M solution off a metallic salt M(NO_(3)) reacted exactly with 25 mL of 0.1 M solution of NaSO_(3) in the reaction SO_(3)^(2-) is oxidized to SO_(4)^(2-) . If in the original salt oxidation number fo the metal was 3, what would be the new oxidation number of metal?

When 1L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide then the molarity of sodium sulphate formed is (H_(2)SO_(4) + 2NaOH rarr Na_(2)SO_(4) + 2H_(2)O) :

Write the balanced equations for the following reactions : * (iv) H_(2)SO_(4_((aq)))+NaOH_((aq))rarrNa_(2)SO_(4_((aq)))+H_(2)O_((1))

The volume of 0.1M H_(2)SO_(4) solution required to neutralise 50ml of 0.2M NaOH solution is -

Consider the reaction: H_(2)SO_(3)(aq)+Sn^(4+)(aq)+H_(2)O(l)toSn^(2+)(aq)+HSO_(4)^(-)(aq)+3H^(+)(aq) Which of the following statements is correct?

________ ml of 0.1 M H_(2)SO_(4) is required to neutralise 50 ml of 0.2 M NaOH solution