Question

A chemical analysis of a surface water yields the following data:

Ion      Concentration (mg/L)

Ca2+ 90

Mg2+ 30

Na+ 72

K+ 6

Cl- 120

SO42- 225

HCO3- 165

pH 7.5

What concentration of Cl- (mg/L) would make the cations and anions balance?  

Answer 1

pH 7.5

H+= 10^-7.5=3.16*10^-8

OH-= 10^6.5=3.16*10^-7

What concentration of Cl- (mg/L) would make the cations and anions balance?  

First calculate the total moles of anion-cation as follows:

Mole of HCO3- = 165.0 mg or 0.165 g / 61.0168 g/mol

=2.70*10^-3 mol

Mole of SO42- = 225 mg or 0.225 g / 96.0626 g/mol

=2.34*10^-3 mol

OH-= 10^6.5=3.16*10^-7

Now total number of negative charge =2.70*10^-3 +(2*2.34*10^-3 )+ 3.16*10^-7

= 0.0073 mol negative charge

moles of Ca2+ = 90 mg or 0.090 g/40.08 g/mol =2.25*10^-4 mol

moles of Mg2+ = 30 mg or 0.030 g/24.31 g/mol =1.23*10^-3 mol

moles of K+ = 6 mg or 0.006 g/39.098 g/mol =1.53*10^-4 mol

H+= 10^-7.5=3.16*10^-8

Now total number of positive charge =(2.25*10^-4)*2+(1.23*10^-3)*2+1.53*10^-4+ 3.16*10^-8

= 0.0031 mol positive

Now calculate the charge by Cl-=

=0.0073 mol negative charge – 0.0031 mol positive

= 0.0042 Mol Cl-

Molar mass of Cl- 35.45 g/mol

Now calculate the amount of Cl= 35.45 g/mol*0.0042 Mol Cl-

= 0.145889 g or 1.45*10^4 mg/L Cl-