Question

Calculate the ionic strength of a solution that is a. 0.040 Min FeSO4. Ionic strength = b. 0.40 M in (NH4), Cr 04. Ionic strength = c. 0.40 Min FeCl2 and 0.40 M in FeCl2. Ionic strength = d. 0.040 M in La (NO3)2 and 0.050 M in Fe(NO3)2- Ionic strength =

Answer 1

Answero- ionic strength = ECZ7 (0) 2 ini. C = concentration a=ion charges. i = ionic strength a) Fesou Fetz + sõja Ooovom O-ovom 0.040 in ſcret" } [+]+[56732-13] iz [co-040)(4) + (0-040)(y)] 1-$1 0.160 + 0-160) 1-460.320) 1 -0160 m

(94,), Cr, Fanni + cord 2x040 0380m 0.40 0-40M i = [[nn.*])** (cm6234] 1- $ [6-801en + (046) (4] iz [ 0.80 + 1-60] 1= [ 2.40] = 1-20m Fecl, O.yom ² Fe 3+ 3ci 0-40m 380-40m 1-2m iss [[Fet"] (+2+ + Oc7][13] 9) + (1-2) (9) 1 - 5( 3-6 +10,8) = $14.4)= logo

Fech 0.40m n Fetz+201 0.40m (280-40)m 0380m i- • [[fety(+2] + [61][-1"] 1 = { [00:46)(w+ Co-fo) (v] izţ ( 106+0.8) is +(2+4) = 1-2m ② Lac No) a hate + 3 NO 1- $ (reats (35€ + [no] 141"} 1-5 [Co-ovo)9) + (0.120) (9) iz (0-360] +0.120) 9-5 (0.480) = 0.22m Ooyon 13X0-040) (0.120m

e Feing)Fet teno ooosom 6.osom (280-0601 m Ooloom P = ([Fety (25++ (wo) 43) *- tf co-oso)4 + (0-100)(0) iz 5(8.20 +00) 1- (0.3) i = 0.15m