Solve this:
Q. 2 M of 100 mL Na2SO4 is mixed with 3 M of 100 mL NaCl solution and 1 M of 200 mL caC12 solution. Then, the ratio of the concentration of cation and anion is
(1) 1:1 (2) 2:1
(3) 2:3 (4) 1:2
Dear student ,
2 M Na2SO4 = 4 M Na+ and 2M SO42- (I)
3 M NaCl = 3M Na+ and 3 M Cl- (II)
and 1 M CaCl2 = 1M Ca2+ and 2 M Cl- (III)
now on mixing the above three solution total vol. = 400 ml
then for cations ; M V = M1V1 + M2V2 + M3V3
mix. (I) (II) (III)
M 400 = 4 100 + 3 100 + 1200
M = 9 /4 (total conc. of cations ) ------(i)
and for anions ; M V = M1V1 + M2V2 + M3V3
mix. (I) (II) (III)
M 400 = 2 100 + 3 100 + 2 200
M = 9 /4 (total conc . of anions ) -----(ii)
from eq. (i) and (ii) ; ratio of the conc. of cations and anions = 1 : 1 (option 1)
Regards
2 M Na2SO4 = 4 M Na+ and 2M SO42- (I)
3 M NaCl = 3M Na+ and 3 M Cl- (II)
and 1 M CaCl2 = 1M Ca2+ and 2 M Cl- (III)
now on mixing the above three solution total vol. = 400 ml
then for cations ; M V = M1V1 + M2V2 + M3V3
mix. (I) (II) (III)
M 400 = 4 100 + 3 100 + 1200
M = 9 /4 (total conc. of cations ) ------(i)
and for anions ; M V = M1V1 + M2V2 + M3V3
mix. (I) (II) (III)
M 400 = 2 100 + 3 100 + 2 200
M = 9 /4 (total conc . of anions ) -----(ii)
from eq. (i) and (ii) ; ratio of the conc. of cations and anions = 1 : 1 (option 1)
Regards