Question

Part A) Three different aqueous solutions are made at 25 °C, each containing one of the amphoteric salts below. Classify each solution as acidic, neutral, or basic. Ionization constants can be found here. (https://sites.google.com/site/chempendix/ionization)

1) sodium hydrogen carbonate

2) lithium hydrogen oxalate

3) potassium dihydogen phosphite

The ionization constants are only to be used in part A

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Part B) Assuming equal concentrations, rank these solutions by pH.

Please order pH from highest to Lowest

NH4NO3 (aq), LiOH (aq), HNO3 (aq), Mg(ClO)2 (aq), CaBr2 (aq)

Answer 1

PART A

1) sodium hydrogen carbonate (NaHCO₃)

When NaHCO₃ is dissolved in water, it dissociates almost completely into Na⁺ and HCO₃^- ions.
NaHCO₃(s) + H₂O(l) Na⁺(aq) + HCO₃^-(aq)
When HCO₃^- behaves as an acid or proton donor, the following equilibrium is attained,
HCO₃^-(aq) + H₂O(l) CO₃^2-(aq) + H₃O⁺ (aq)  
For above equilibrium, acid dissociation constant K_a is same as the second ionization constant of H₂CO₃ which is given in the table as, K_a2 = 4.7 * 10^-11.
Hence, acid dissociation constant of HCO₃^- is K_a = 4.7 * 10^-11     ............(1)
When HCO₃^- behaves as a base or proton acceptor, the following equilibrium is attained,
HCO₃^-(aq) + H₂O(l) H₂CO₃(aq) + OH^-(aq)  
For above equilibrium, base dissociation constant K_b can be calculated as shown below. H₂CO₃ and HCO₃^- are conjugate acid-base pair (differ by one H⁺). For conjugate acid-base pair, K_a1 * K_b = 10^-14 where K_a1 and K_b are acid dissociation constant and base dissociation constant of the acid and base respectively.    Hence, K_b = 10^-14 / K_a1 = 10^-14 / (4.5 * 10^-7) = 2.2 * 10^-8 ​    { Since, HCO₃^- can be obtained by first ionization of H₂CO₃, the first ionization constant of H₂CO₃ (K_a1) is considered as acid dissociation constant of H₂CO₃ in above equation. H₂CO₃ + H₂O <------> H₃O⁺ + HCO₃^-      K_a1 = 4.5 * 10^-7}

Hence, base dissociation constant of HCO₃^- is K_b = 2.2 * 10^-8   ..........(2)
From results (1) and (2), it is clear that,
Base dissociation constant of HCO₃^- (K_b) > Acid dissociation constant of HCO₃^- (K_a)
Hence, aqueous solution of sodium hydrogen carbonate (NaHCO₃) is basic in nature.

2) lithium hydrogen oxalate (Li^+-OOC-COOH or LiHC₂O₄)
When LiHC₂O₄ is dissolved in water, it dissociates almost completely into Li+ and HC₂O₄^- ions.
LiHC₂O₄ (s) Li⁺(aq) + ^-OOC-COOH(aq) When HC₂O₄^- behaves as an acid or proton donor, the following equilibrium is attained, HC₂O₄^- (aq) + H₂O(l) C₂O₄^2-(aq) + H₃O⁺ (aq) For above equilibrium, acid dissociation constant K_a is same as the second ionization constant of H₂C₂O₄ which is given in the table as, K_a2 = 1.5 * 10^-4.    Hence, acid dissociation constant of HC₂O₄^- is K_a = 1.5 * 10^-4     ............(3) When HC₂O₄^- behaves as a base or proton acceptor, the following equilibrium is attained,    HC₂O₄^- (aq) + H₂O(l) H₂C₂O₄ (aq) + OH^-(aq)    For above equilibrium, base dissociation constant K_b can be calculated as shown below. H₂C₂O₄ and HC₂O₄^- are conjugate acid-base pair (differ by one H⁺). For conjugate acid-base pair, K_a1 * K_b = 10^-14 where K_a1 and K_b are acid dissociation constant and base dissociation constant of the acid and base respectively. Hence, K_b = 10^-14 / K_a1 = 10^-14 / (5.6 * 10^-2) = 1.8 * 10^-13    { Since, HC₂O₄^- can be obtained by first ionization of H₂C₂O₄, the first ionization constant of H₂C₂O₄ (K_a1) is considered as acid dissociation constant of H₂C₂O₄ in above equation. H₂C₂O₄ + H₂O H₃O⁺ + HC₂O₄^-       K_a1 = 5.6 * 10^-2} Hence, base dissociation constant of HC₂O₄^- is K_b = 1.8 * 10^-13   ..........(4)    From results (3) and (4), it is clear that, Acid dissociation constant of HC₂O₄^- (K_a) > Base dissociation constant of HC₂O₄^- (K_b) Hence, aqueous solution of lithium hydrogen oxalate (LiHC₂O₄) is acidic in nature.

3) potassium dihydrogen phosphite (KH₂PO₃)
When KH₂PO₃ is dissolved in water, it dissociates almost completely into K⁺ and H₂PO₃^- ions.
KH₂PO₃(s) + H₂O(l) K⁺(aq) + H₂PO₃^- (aq)
When H₂PO₃^- behaves as an acid or proton donor, the following equilibrium is attained,
H₂PO₃^- (aq) + H₂O(l) HPO₃^2-(aq) + H₃O⁺ (aq)  
For above equilibrium, acid dissociation constant K_a is same as the second ionization constant of H₃PO₃ (phosphorous acid) which is given in the table as, K_a2 = 2 * 10^-7.
Hence, acid dissociation constant of H₂PO₃^- is K_a = 2 * 10^-7     ............(5)
When H₂PO₃^- behaves as a base or proton acceptor, the following equilibrium is attained,
H₂PO₃^- (aq) + H₂O(l) H₃PO₃(aq) + OH^-(aq)  
For above equilibrium, base dissociation constant K_b can be calculated as shown below.
H₃PO₃ and H₂PO₃^- are conjugate acid-base pair (differ by one H⁺). For conjugate acid-base pair, K_a1 * K_b = 10^-14 where K_a1 and K_b are acid dissociation constant and base dissociation constant of the acid and base respectively.
Hence, K_b = 10^-14 / K_a1 = 10^-14 / (5 * 10^-2) = 2 * 10^-13
{ Since, H₂PO₃^- can be obtained by first ionization of H₃PO₃, the first ionization constant of H₃PO₃ (K_a1) is considered as acid dissociation constant of H₃PO₃ in above equation.
H₃PO₃ + H₂O H₃O⁺ + H₂PO₃^-       K_a1 = 5 * 10^-2}
Hence, base dissociation constant of H₂PO₃^- is K_b = 2 * 10^-13   ..........(6)
From results (5) and (6), it is clear that,
Acid dissociation constant of H₂PO₃^- (K_a) > Base dissociation constant of H₂PO₃^- (K_b)
Hence, aqueous solution of potassium dihydrogen phosphite (KH₂PO₃) is acidic in nature.

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PART B
ANSWER: LiOH(aq) > Mg(ClO)₂(aq) > CaBr₂(aq) > NH₄NO₃(aq) > HNO₃(aq)
Explanation:    pH is higher for less acidic or more basic solution and it is lower for more acidic or less basic solution. For neutral solution pH is 7.

LiOH is a strong lewis base that dissociates almost completely and furnishes OH^- ions in aqueous solution. HNO₃ is a strong lewis acid which dissociates almost completely and furnishes H⁺ ions (to form H₃O⁺ )in aqueous solution. Hence, pH is highest for LiOH(aq) and lowest for HNO₃(aq).    LiOH(s) Li⁺(aq) + OH^-(aq) HNO₃ + H₂O H₃O⁺ + NO₃^-

CaBr₂(aq) : CaBr₂ is salt of strong acid (HBr) and strong base {Ca(OH)₂} and hence CaBr₂ (aq) will be neutral or its pH will be 7.

In general, conjugate acid of a weak base is stronger while that of a strong base will be weaker. Similarly, conjugate base of weak acid is stronger while that of strong acid will be weaker.

NH₄NO₃(aq): NH₄NO₃ dissociates into NH₄⁺ and NO₃^- ions in aqueous solution. NH₄NO₃(s)​ NH₄⁺(aq) + NO₃^-(aq)    Since nitric acid is a strong acid, its conjugate base NO₃^- is a very weak base. NH₄⁺ and NH₃ are a conjugate acid-base pair. H₂O and OH^- are also a conjugate acid-base pair. Since NH₃ is a base much weaker than OH^-, NH₄⁺ is stronger acid than H₂O. Hence, NH₄⁺ can donate proton to H₂O leading to formation of H₃O⁺ ions.    NH₄⁺ + H₂O NH₃ + H₃O⁺    NH₃ is a weak base and the formation of H₃O⁺ makes the solution acidic.    Hence, NH₄NO₃(aq) is acidic and its pH will be lower than 7.

Mg(ClO)₂ (aq): Mg(ClO)₂ dissociates almost completely into Mg²⁺ and ClO^- ions in aqueous solution.    Mg(ClO)₂ ​(s) Mg²⁺(aq) + 2ClO^- (aq)    HClO and ClO^- are a conjugate acid-base pair. H₃O⁺ and H₂O are also a conjugate acid-base pair. Since, HClO is weak acid than H₃O⁺, ClO^- is a stronger base than H₂O. Hence, ClO^- can abstract H⁺ from H₂O leading to formation of OH^- ions.
ClO^- + H₂O HClO + OH^-
HClO is a weak acid and formation of OH^- ions makes the solution basic. Hence, Mg(ClO)₂ (aq) is basic and its pH will be greater than 7.

Hence, given solutions can be placed in order of their pH (from highest pH to lowest pH) as shown below. LiOH(aq) > Mg(ClO)₂(aq) > CaBr₂(aq) > NH₄NO₃(aq) > HNO₃(aq)