Question

Chapter is over Equilibrium Chemistry and connecting to Gravimetric systematic solving.

6. Write a charge balance equation and mass balance equations for the following solutions. Some solutions may have more than one mass balance

Please help I don't understand how to derive these.

a. 0.1 M NaCL

b. 0.1 M HCl

c. 0.1 M HF

g. 0.10 M HCl and 0.05 M NaNO2

Answer 1

a.

NaCl (strong electrolyte) is found in solution as Na⁺_(aq) 0.1 M and Cl^-_(aq) 0.1 M. No one react with water.

Mass balance :

C_NaCl (formal) = ([Na+] + [Cl-])/2 or

NaCl mass = mass of Na+   + mass of Cl-

Charge balance:

(+1)[Na+] + (-1)[Cl-] = 0

b.

HCl is a strong acid. It react completely with water.

HCl + H2O = H3O+   + Cl-

Mass balance :

C_HCl (formal) = ([H3O+] + [Cl-])/2

Charge balance:

(+1)[H3O+] + (-1)[Cl-] = 0

You can add also the equilibrium

2H2O = H3O+   + Cl-

And write

(+1)[H3O+]_total + (-1)[Cl-]   + (-1)[HO-] = 0

Or [H3O+]_total = [Cl-]   + [HO-]       (but [HO-] is only 1x10^-13M, a negligible value)

c.

HF is a weak acid (Ka = 7.2x10^-4 , pKa = 3.14)

[H+] = [F-] = (Ka.Cacid)^1/2 = 8.5 x10^-3M

Mass balance :

C_HF (formal) = [HF] + ([F-] + [H+])/2

Charge balance:

[H+] = [F-]

Or considering also the dissociation of H2O

[H+]_total = [F-] +[HO-]

d.

NaNO2 (strong electrolyte) is found in solution as Na⁺_(aq) 0.1 M and NO₂^-_(aq) 0.1 M. NO₂^- react with H⁺ (from HCl) and is completely neutralized. So the solution will contains:

Na⁺_(aq) 0.1 M and Cl^-_(aq) 0.1 M and the weak acid HNO2 that dissociates (see b.)

Mass balance :

C_NaNO2 + C_HCl = ([Na+] + [Cl-] ) /2 + [HNO2] + [NO2-]

Charge balance

[H+] + [Na+] = [Cl-] + [NO2-]

Or considering also the dissociation of H2O

[H+]_total + [Na+] = [Cl-] + [NO2-] + [HO-]