Net Ionic Equations for Aqueous Solutions ( continued ) Illustrate with a net io
ID: 704377 • Letter: N
Question
Net Ionic Equations for Aqueous Solutions (continued)
Illustrate with a net ionic equation for each of the following how the pH would increase, decrease, or stay the same with the indicated changes:
(b) Adding ammonium nitrate to a solution of ammonium hydroxide
(c) Adding hydrochloric acid to a solution of ammonia
(d) Adding potassium formate to a solution of formic acid
(e) Adding sodium hypobromite to a solution of hypobromous acid
(f) Adding potassium perchlorate to a solution of potassium hydroxide
Explanation / Answer
b) The ionization of ammonium hydroxide (NH4OH) in aqueous solution is given as
NH4OH (aq) --------> NH4+ (aq) + OH- (aq)
Since OH- is formed, we define the Kb (base ionization constant) as
Kb = [NH4+][OH-]/[NH4OH] …..(1)
NH4NO3 ionizes completely in aqueous solution as
NH4NO3 (aq) --------> NH4+ (aq) + NO3- (aq)
NO3- is spectator ion (doesn’t participate in the reaction); however, [NH4+] increases and thus the numerator in expression (1) tends to increase. However, Kb is thermodynamic equilibrium constant and must remain constant at a particular temperature. Thus, to counter-balance the effect of increased [NH4+], [NH4OH] must increase proportionately. This is possible only when [OH-] decreases, i.e, the ionization of NH4OH is suppressed. Thus, [OH-] falls and pOH = -log [OH-]
increases.
It is known that pH + pOH = 14. Since pOH increases, the pH must decrease, i.e, the solution becomes more acidic.
c) Consider the ionization of ammonia (NH3) in water.
NH3 (aq) + H2O (l) -------> NH4+ (aq) + OH- (aq)
Since OH- is formed, we define the Kb (base ionization constant) as
Kb = [NH4+][OH-]/[NH3] …..(1)
HCl ionizes completely in aqueous solutions as
HCl (aq) -------> H+ (aq) + Cl- (aq)
Cl- is the spectator ion. However, H+ combines with OH- to form water (H2O) which remains undissociated.
H+ (aq) + OH- (aq) -------> H2O (l)
Due to the reaction with water, [OH-] decreases. Therefore, pOH = -log [OH-] increases and hence, pH = 14 – pOH decreases. Thus, the solution becomes more acidic.
d) Consider the ionization of formic acid (HCOOH) as below.
HCOOH (aq) --------> H+ (aq) + HCOO- (aq)
Since H+ is produced, we define the acid ionization constant
Ka = [H+][COO-]/[HCOOH]
Potassium formation (K+HCOO-) ionizes completely in aqueous solution as
K+HCOO- (aq) --------> K+ (aq) + HCOO- (aq)
K+ is the spectator ion. However, [HCOO-] increases. But Ka must remain constant at a particular temperature. To keep Ka constant, [H+] must decrease. A decrease in [H+] leads to an increase in pH since pH = -log [H+].
e) The ionization of hypobromous acid (HOBr) is given as
HOBr (aq) -------> H+ (aq) + BrO- (aq)
Since H+ is produced, we define the acid ionization constant
Ka = [H+][BrO-]/[HOBr]
Sodium hypobromite (NaOBr) ionizes completely in aqueous solution as
NaOBr (aq) --------> Na+ (aq) + BrO- (aq)
Na+ is the spectator ion. However, [BrO-] increases. But Ka, being a thermodynamic equilibrium constant, must remain constant at a particular temperature. To keep Ka constant, [H+] must decrease. A decrease in [H+] leads to an increase in pH since pH = -log [H+].
f) Potassium hydroxide (KOH) ionizes as
KOH (aq) --------> K+ (aq) + OH- (aq)
KOH is a strong base and ionizes completely.
Potassium perchlorate (KClO4) is the salt of a strong base (KOH) and a strong acid (perchloric acid, HClO4) and ionizes completely to produce ions.
KClO4 (aq) -------> K+ (aq) + ClO4- (aq)
None of the ions can participate in acid-base equilibrium and thus, there is no change in [OH-]. Thus, the pOH and the pH of the solution remain unchanged. Infact, the pH of a solution of a strong acid or base is much less influenced by the addition of salts.