What is the concentration of Sulfate ion in a 0.5 M solution of Al2(SO4)3?
3 answers
0.5M x (3 SO4^2- ions/1 Al2SO4)3 molecule) = 0.5 x 3/1 = ?
ygc
Your Answer Will Be 3N Because,
N=n*M (Where N = Normality, n = Acidity, Basicity or Neutrality , M=Molarity)
•°•n=Neutrality Of Al2(SO4)3
=2Al³+ + 3SO4²-
n=6
•°•N=6*0.5(Which Is Molarity Of Aqueous)
=3.0N
THANK YOU 🙏✌️💜
N=n*M (Where N = Normality, n = Acidity, Basicity or Neutrality , M=Molarity)
•°•n=Neutrality Of Al2(SO4)3
=2Al³+ + 3SO4²-
n=6
•°•N=6*0.5(Which Is Molarity Of Aqueous)
=3.0N
THANK YOU 🙏✌️💜
1 answer