Question

110mg CaCl2 and 50mg CaSO4 added to 500mL water. solution pH is 8.0 ([H+]=10E-8M ; [OH-]= 10E-6 M). assuming all salts dissociate completely express concentrstion of Ca2+, Cl-, SO42- in mg/L, millimolar, normal (N) units. if total molar concentrations of all speicies in solution is same as in pure water, what are mole fractions of H+, OH-, Ca2+, Cl-, SO42-, H2O

Answer 1

Your question is very long and its time consuming.

I have solved 1st 3 parts of it:

Molar mass of CaCl2 = 111 gm
Number of moles of CaCl2 = 110*10^-3/111 = 9.91*10^-4 mol
[CaCl2] = 9.91*10^-4 mol / 0.5L = 4.96*10^-4 mol/L
[Cacl2] = 110 mg/0.5 L = 220 mg/L

Molar mass of CaSO4 = 136 gm
Number of moles of CaSO4 = 50*10^-3/136 = 3.68*10^-4 mol
[CaSO4 ] = 3.68*10^-4 mol / 0.5 L = 1.84*10^-4 mol/L
[CaSO4 ] = 50 mg/0.5 L = 100 mg/L

Concentration of Ca2+: = 4.96*10^-4 + 3.68*10^-4 = 8.64*10^-4 mol/L
                                   =0.864 millimolar
                                   = 0.864*40 =34.56 mg/L {since molar mass of Ca= 40 gm}
                                   =0.864/2 N =0.432 N

Concentration of Cl- = 2* Concentration of Ca2+: from CaCl2
                               =9.92*10^-4 mol/L
                                =0.992 millimolar
                               = 0.992*35.5 = 35.22 mg/L {since molar mass of Ca= 35.5 gm}
                               = 0.992 N

Concentration of so42- = Concentration of CaSO4
                               = 1.84*10^-4 mol/L
                                =0.184 millimolar
                               = 0.184*96 = 17.66 mg/L {since molar mass of SO4= 96 gm}
                               = 0.184/2 = 0.092 N