Chemical equilibrium question in image attached answer it fast
mayank nasa , 7 Years ago
Grade 11
Physical Chemistry> Chemical equilibrium question in image at...
2 Answers
Nishant Sharma
Last Activity: 7 Years ago
in this first we calculate the conc. of Ag+ by the information given in second half of the ques. Since in acidic medium the KMnO4 acts as an oxidising agent with z factor 5 wich involved the reaxn. as shown
MnO4- + 8H+ + 5e- ----> Mn+2 + 4H2O
therefore from equvalance concept
N1V1 = N2V2 where N1 is the normality of Ag+ and V1 is the vol. of its sol. and N2 and V2 are the normality and vol. of KMnO4 sol. respectively
i.e., N1*25 = 5*30*.0832 => N1 =
Nishant Sharma
Last Activity: 7 Years ago
in this first we calculate the conc. of Ag+ by the information given in second half of the ques. Since in acidic medium the KMnO4 acts as an oxidising agent with z factor 5 wich involved the reaxn. as shown
MnO4- + 8H+ + 5e- ----> Mn+2 + 4H2O
therefore from equvalance concept
N1V1 = N2V2 where N1 is the normality of Fe+2 and V1 is the vol. of its sol. and N2 and V2 are the normality and vol. of KMnO4 sol. respectively
i.e., N1*25 = 5*30*.0832 => N1 = .4992 N
since the z factor of Fe+2+ is 1 therefore N = M
hence conc. of [Fe+] is 0.4992M
now after mixing of equal vol. of AgNo3 sol. and Fe+2 sol. the new morality of each sol. becomes half
now since the conc. of Fe+2 in sol. is 0.4992M the conc. of Fe+3 will be = (1.09/2 – .04992) = 0.0458M
and the acc. to reaxn. the moles of Fe+3 formed is eual to the moles of Ag+ consumed
therefore from the equvalance concept the conc. of Ag+ becomes (.15 – .0458) = .0292M
now the equilibrium constant is
Kc = [Fe+3]/([Fe+2]*[Ag+])
= .0458/(.0292*.4992)
=3.14M-1L Ans.
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