A sample of water has the following concentrations of ions (and pH = 7.0): catio

Question

A sample of water has the following concentrations of ions (and pH = 7.0):

                        cations                                    mg/L      anions            mg/L

                        Ca+2      40.0                           HCO3-   110.0

                        Mg+2     10.0                           SO42-     67.2

                        Na+       ?                                 Cl-         11.0

                        K+         7.0

Assuming no other constituents are missing, use an anion-cation balance to estimate the concentration of Na+ (in mg/L as Na+)? Remember that the balance cannot be in mg/L.

Explanation / Answer

See here pH = 7.0.

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

Mole of HCO3- = 110.0 mg or 0.110 g / 61.0168 g/mol

=1.8*10^-3 mol

Mole of SO42- = 67.2 mg or 0.0672 g / 96.0626 g/mol

=0.70 mol

Mole of Cl- = 11.0 mg or 0.011 g / 35.45 g/mol

=3.10*10^4 mol

Now total number of negative charge =1.8*10^-3 mol +2*0.70 mol+3.10*10^4 mol

= 1.402 mol negative charge

moles of Ca2+ = 40 mg or 0.040 g/40.08 g/mol =9.98*10^-4 mol

moles of Mg2+ = 10 mg or 0.010 g/24.31 g/mol =4.11*10^-4 mol

moles of K+ = 7 mg or 0.007 g/39.098 g/mol =1.79*10^-4 mol

Now total number of positive charge =9.98*10^-4 mol*2+4.11*10^-4 mol*2+1.79*10^-4

= 2.997*10^-3 mol positive

Now calculate the charge by Na+=

=1.402 mol negative charge - 2.997*10^-3 mol positive

= 1.3999 Mol Na+

Molar mass of Na 22.9898 g/mol

Now calculate the amount of Na = 22.9898 g/mol*1.3999 Mol Na+

= 32.163 g or 3.2*10^4 mg/L

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