Charge balance equations (chemistry)

Charge balance equations are derived using the concept of electroneutrality, where the sum of positive charges equals to the sum of negative charges in a solution. Such equations are useful for analysing acid-base equilibria and formulating complex acid-base titration equations.

Consider a solution containing water, a strong acid of concentration Ca, and a strong base of concentration Cb, with the following equilibria:

H_2O(l)\rightleftharpoons H^+(aq)+OH^-(aq)

HA(aq)\rightleftharpoons H^+(aq)+A^-(aq)

BOH(aq)\rightleftharpoons B^+(aq)+OH^-(aq)

To maintain electroneutrality, the sum of the number of moles of cations H+ and B+ must equal to that of anions A and OH. As the volume of the solution is common to all ions,

[H^+]+[B^+]=[OH^-]+[A^-]

When formulating charge balance equations for aqueous compounds with multiple equilibria, we need to account for every charged species on the LHS of a particular equilibrium, which can be complicated. To avoid mistakes,  we select equilibrium expressions where species on the LHS are neutral. For example, the equilibrium equations for a diprotic acid can be presented in the following ways:

H_2A(aq)\rightleftharpoons H^+(aq)+HA^-(aq)

HA^-(aq)\rightleftharpoons H^+(aq)+A^{2-}(aq)

H_2A(aq)\rightleftharpoons 2H^+(aq)+A^{2-}(aq)

Select the first and third equilibria, i.e., we can imagine part of the initial number of moles of H2A dissociating into Ha+ and HA, with the remaining part of the initial number of moles of H2A dissociating into 2Hb+ and A2-. For the first equilibrium, the charge balance equation is

[H_a^{\, +}]=[HA^-]\; \; \; \; \; \; \; \; (31)

For the third equilibrium, the charge balance equation is

[H_b^{\, +}]=2[A^{2-}]\; \; \; \; \; \; \; \; (32)

Combining eq31 and eq32,

[H_a^{\, +}]+[H_b^{\, +}]=[HA^-]+2[A^{2-}]

[H^+]=[HA^-]+2[A^{2-}]

 

Question

Write the charge balance equation for a triprotic acid.

Answer

Since,

H_3A(aq)\rightleftharpoons H^+(aq)+H_2A^-(aq)

H_3A(aq)\rightleftharpoons 2H^+(aq)+HA^{2-}(aq)

H_3A(aq)\rightleftharpoons 3H^+(aq)+A^{3-}(aq)

we have,

[H^+]=[H_2A^-]+2[HA^{2-}]+3[A^{3-}]

 

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