Chemical equivalents

ISC class 11_unit 1-Basic concepts

1. Lets watch this video and try to get the central idea, Todays lesson is about this central idea https://www.youtube.com/watch?v=9OEf5IYsRi8 2 min

2. Chemical Equivalents https://www.youtube.com/watch?v=WCr5UVf-vKM Till 2.14 min

3. Equivalent weight or equivalent mass Equivalent weight of the given substance is the number of parts by mass of the given substance that displaces or combines with 1.008 parts H 2 ; 8 parts oxygen and 35.5 parts by mass of chlorine. ** We know that in any chemical reaction definite mass of one substance reacts with particular mass of other substance. Ex – in water one part of H reacts with 16 part of O. That’s mean- “In a chemical reaction a definite mass of one substance is always equivalent to definite mass of other substance this specific mass is called as its equivalent mass.”

4. https://www.youtube.com/watch?v=EIc9ZbKKDxE Lets understand what is equivalent weight with the help of this video Watch till 3.40 min

5. Zn + H2SO4 ----  ZnSO4 + H2 65 98 2 2 parts by mass of hydrogen = 65 parts by mass of Zn ------------”-------------------- = 65 /2 = 32.5 Similarly 2 parts by mass of hydrogen = 98 parts by mass of H2SO4 1 --------------- “ ----------------- = 98/2 = 49 * eq. weight is just a number it does not posses any unit Equivalent weight can be worked out experimentally also.

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9. Equivalent weight of an element Atomic mass of an element Eq. Wt. of an element= Valency [Valency is the combining capacity of any substance] Example:- Eq. Wt of oxygen= 16/2= 8 Eq. wt of Mg= 24/2= 12 Eq wt of iodine = 127/1=127 Eq. wt of Cu in cupric salt = 63.6/2=31.8 Eq. wt of Cu in cuprous salt = 63.6/1=63.6 Variable valency effects eq wt.

10. Find Eq wt of following Al= 27 Fe= 56 Al Fe 2+ Fe 3+

11. Equivalent weight of a normal salt Formula weight of salt Eq. Wt of a normal salt = Total charge on cationic part Example:- Eq. Wt of NaCl = 23 + 35.5 /1 = 58.5 Eq. wt of Na2CO3 = (2 x 23) + 12 + (3 x 16 ) /2= 53

12. Equivalent weight of a Acid salt Formula weight of salt Eq. Wt of a acid salt = number of replaceable H atoms Example:- Eq. wt of NaHCO 3 = (1 x 23) + 1 + 12 + (16 x 3 ) /2 = 84

13. Equivalent weight of an Acid Molecular weight of an acid Eq. Wt of an acid = Basicity of the acid Example:- Eq. wt of H 2 SO 4 = (1 x 2) + 32 + (16 x 4 ) /2 = 98/2= 49 Eq. wt of HCl = 1 +35.5/1= 36.5/1 = 36.5

14. Equivalent weight of a base Molecular weight of the base Eq. Wt of a base = Acidity of the base Example:- Eq. wt of Ca(OH) 2 = 40 + 2(16 + 1 ) /2 = 74/2= 37 Eq. wt of NaOH = 23 + 16 + 1 = 40/ 1 = 40

15. Equivalent weight of a Oxidant (OA) or reductant (RA) Formula mass Eq. Wt of a OA / RA = number of e- gained or lost by one molecule 126/2=63

16. Gram equivalent weight The mass of a substance in gm, numerically equal to its equivalent weight is called gm equivalent weight or one gm equivalent of that substance. mass of the substance in gm Number of gm equivalents = equivalent weight Substance Mass taken Eq. wt. No. of gm. Eq. H2SO4 4.9g 49 4.9/49=1 H2C2O4 10g 63 10/63=0.16 K2Cr2O7 2g 49 2/49=0.041 NaOH 0.4g 40 0.4/40=0.01

17. Lets understand this little more by connecting with our previous knowledge 2NaOH + H2SO4 ---  Na2SO4 + 2H2O In case of mole we must have a balanced chemical equation to find of how much of NaOH is reacting with How much H2SO4. Therefore came concept of eq wt and using that we can easily Find out that how much of substance A is reacting with how much substance B.

18. Molecular wt and eq wt share similar relations In case of Molecular wt Mass No of mole = Mol. wt In case of equivalent wt Mass No of gm equivalent = eq. wt Mass No of gm eq.= eq wt Mass of sub Mass No of gm eq.= or x X Molecular wt / X Mol. Wt No of gm eq = Mole x X [ X = valency/acidity/ bascity / charge on cation/ no of e- gained or lost] Not in reduced syllabus

19. 2NaOH + H2SO4 ---  Na2SO4 + 2H2O Lets find out no of gm eq when no of moles are given 2 mole x 1 1 mole x 2 No of gm eq = Mole x X [ X will have different values depending upon substance] 2 equivalent 2 equivalent Therefore it is said that in a chemical reaction I equivalent of a substance reacts with 1 eq of other substance.

20. Lets find out no of gm eq when no of moles are given N2 + 3H2 ---  2NH3 No of gm eq = Mole x X [ X will have different values depending upon substance] 1 mole x 3 3 mole x 1 3 gm equivalent 3gm equivalent

21. Relationship between Gm Eq. Wt. (E), Gm Atomic weight (A) and valency(V) A E = V A = E X V Gm atomic mass = Gm eq. wt X valency

22. https://www.youtube.com/watch?v=qGAlv7Gu5s0 Watch this video to revise all the concepts that we have learned today

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35. Find equivalent weight of following CaCO3 HNO3 Page 36; Q = 1.53 Formula wt Valency on cation 100 = 50 2 Formula wt Basicity of acid 63 = 63 1

36. Ca(OH)2 Br Formula wt Acidity of base 74 = 37 2 Atomic mass Valency 80 = 80 1

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