Question

2. (20 pts.) What are the activities (effective concentrations) of ALL the ionic species for the water sample represented in the following table? Constituents mg/L Cations Sodium 16,470 Potassium 120 Calcium 2010 Magnesium 885 Barium 45 Strontium 135 Total iron Dissolved iron Anions Chloride 31,730 Sulphate L 0.5 Bicarbonate

Answer 1

Activity or “effective concentration” : [a] = γ *(C)

γ = activity coefficient ; C = concentration = mol / L

Ionic strength :  I = concentration of ions and charge in solution ; I = 1/2 ΣCi zi ²

First convert given concentration into molality/ molarity from mg/L

Ionic strength for this solution : I = 1/2 (1.9682) = 0.984 mol/L

(molality of respective ions can be seen below )

Using Debye-Hückel Equation, logγi = −Azi ² (I)^1/2 ( for water A= 0.5 L^1/2/mol^1/2 )

• Sodium C = 0.716 M ,  logγi = −A zi ² (I)^1/2 = - 0.5*1*(0.984)^1/2

γ = 0.322 ; so, [a] =  0.322*0.716 M = 0.231 M

• K C = 0.003 M ,  logγi = −A zi ² (I)^1/2 = - 0.5*1*(0.984)^1/2

γ = 0.322 ; so, [a] =  0.322*0.003 M =9.7*10^-4 M

• Ca C = 0.05 M ,  logγi = −A zi ² (I)^1/2 = - 0.5*4*(0.984)^1/2

γ = 0.0104 ; so, [a] =  0.0104*0.05 M = 5.2*10^-4 M

• Mg   C = 0.036 M ,  logγi = −A zi ² (I)^1/2 = - 0.5*4*(0.984)^1/2

γ = 0.0104 ; so, [a] =  0.0104*0.036 M = 3.7*10^-4 M

• Ba C = 3.3*10^-4 M ,  logγi = −A zi ² (I)^1/2 = - 0.5*4*(0.984)^1/2

γ = 0.0104 ; so, [a] =  0.0104​​​​​​​*3.3*10^-4 M = 3.4*10^-6 M

• Sr C = 1.5*10^-3 M ,  logγi = −A zi ² (I)^1/2 = - 0.5*4*(0.984)^1/2

γ = 0.0104 ; so, [a] =  0.0104​​​​​​​*1.5*10^-3 M = 1.6*10^-5 M

• Total iron C = 2.7*10^-4 M ,  logγi = −A zi ² (I)^1/2 = - 0.5*4*(0.984)^1/2

γ = 0.0104 ; so, [a] =  0.0104​​​​​​​* 2.7*10^-4 M   = 2.8*10^-6 M

​​​​​​​

• Dissolved iron C = 5.27*10^-5 M ,  logγi = −A zi ² (I)^1/2 = - 0.5*4*(0.984)^1/2

γ = 0.0104 ; so, [a] =  0.0104​​​​​​​* 5.27*10^-5 M   = 5.5*10^-7 M

• Cl    C = 0.894 M ,  logγi = −A zi ² (I)^1/2 = - 0.5*1*(0.984)^1/2

γ = 0.322 ; so, [a] =  0.322*0.894 M =0.288 M

• Sulfate C = 1.7*10^-5 M ,  logγi = −A zi ² (I)^1/2 = - 0.5*4*(0.984)^1/2

γ = 0.0104 ; so, [a] =  0.0104​​​​​​​*1.7*10^-5 M = 1.8*10^-7 M

• Bicarbonate   C = 1.2*10^-3 M ,  logγi = −A zi ² (I)^1/2 = - 0.5*1*(0.984)^1/2

γ = 0.322 ; so, [a] =  0.322*1.2*10^-3 M = 3.8*10^-4 M

​​​​​​​