Question

List the species present in an aqueous solution of each of the following:

a. HF b. KOH c. NH3 d. HCl e. Ba(OH)2

Given the concentrations of the following solutions provide the pH, pOH, H3O, and OH-

a. 0.0025 M HCl b. 0.0025 M NaOH c. 0.035 M Ba(OH)2 d. 0.0015 M H2SO4

Answer 1

a) HF is a molecular compound. It is a weak electrolyte. It is not completely dissociated into water.

HF (aq) + H₂O (l)   H₃O ⁺ (aq) + F ^- (aq)

Water undergoes auto ionization as 2 H₂O (l)   H₃O ⁺ (aq) + OH ^- (aq)

Hence, Species present in the aqueous solution of HF are, HF , H₃O  ⁺, F ^- , H₂O and OH ^-

b) KOH is a strong electrolyte. It dissociates completely into water.

KOH (aq) K ⁺ (aq) + OH ^- (aq)

Water undergoes auto ionization as 2 H₂O (l)   H₃O ⁺ (aq) + OH ^- (aq)

Hence, Species present in the aqueous solution of KOH are, K ⁺ , H₃O ⁺,OH ^- and H₂O

c) Ammonia ( NH₃)

It is a weak electrolyte. It is not completely dissociated into water.

NH₃ (aq) + H₂O (l)   NH₄⁺ (aq) + OH ^- (aq)

Water undergoes auto ionization as 2 H₂O (l)   H₃O ⁺ (aq) + OH ^- (aq)

Hence, Species present in the aqueous solution of NH₃ are NH₃ , H₃O ⁺,NH₄⁺ , H₂O and OH ^-

d) HCl

HCl is a strong electrolyte. It dissociates completely into water.

HCl (aq) + H₂O (l)    H₃O ⁺ (aq) + Cl ^- (aq)

Water undergoes auto ionization as 2 H₂O (l)   H₃O ⁺ (aq) + OH ^- (aq)

Hence, Species present in the aqueous solution of HCl are Cl ^- , H₃O ⁺,OH ^- and H₂O

e) Ba(OH) ₂

Ba(OH) ₂ is a strong electrolyte. It dissociates completely into water

Ba(OH) ₂ (aq) Ba ²⁺(aq) + 2 OH ^- (aq)

Water undergoes auto ionization as 2 H₂O (l)   H₃O ⁺ (aq) + OH ^- (aq)

Hence, Species present in the aqueous solution of Ba(OH) ₂ are Ba ²⁺ , H₃O ⁺,OH ^- and H₂O

a) 0.0025 M HCl

We have, pH = - log [H₃O ⁺ ]

Therefore, pH = - log 0.0025 = 2.60

We have, pH + pOH =14

pOH = 14 - pH = 14-2.60 = 11.4

We have, [H₃O ⁺ ] [OH ^-]= 10 ^-14

Therefore, [OH ^-]= 10 ^-14 / 0.0025 = 4.0 10 ^-12 M

ANSWER : pH = 2.60 , pOH = 11.4 , [H₃O ⁺ ] = 0.0025 M , [OH ^-] = 4.0 10 ^-12 M

b) 0.0025 M NaOH

We have, pOH = - log [OH ^-] = - log 0.0025 = 2.60

We have, pH + pOH =14

pH = 14 - pOH = 14-2.60 = 11.4

We have, [H₃O ⁺ ] [OH ^-]= 10 ^-14

Therefore, [H₃O ⁺ ] = 10 ^-14 / 0.0025 = 4.0 10 ^-12 M

ANSWER : pH = 11.4 , pOH = 2.60 , [H₃O ⁺ ] =4.0 10 ^-12 M , [OH ^-] = 0.0025 M

C) 0.035 M Ba(OH) ₂

Consider dissociation of Ba(OH) ₂ in water. Ba(OH) ₂ (aq) Ba ²⁺ (aq) + 2 OH ^- (aq)

From above reaction, [OH ^-] = 2 [Ba(OH) ₂ ] = 2 * 0.035 = 0.070 M

We have, pOH = - log [OH ^-] = - log 0.070 = 1.15

We have, pH + pOH =14

pH = 14 - pOH = 14-1.15= 12.85

We have, [H₃O ⁺ ] [OH ^-]= 10 ^-14

Therefore, [H₃O ⁺ ] = 10 ^-14 / 0.070 = 1.43 10 ^-13 M

ANSWER : pH = 12.85 , pOH = 1.15 , [H₃O ⁺ ] =1.43 10 ^-13 M , [OH ^-] = 0.070 M

D) 0.0015 M H₂SO₄  

Consider dissociation of H₂SO₄ in water. H₂SO₄ (aq) + H₂O (l) 2 H₃O ⁺ (aq) + SO ₄^2- (aq)

From above reaction,  [H₃O ⁺ ] = 2 [H₂SO₄] = 2* 0.0015 = 0.003 M

We have, pH = - log [H₃O ⁺ ]

Therefore, pH = - log 0.003 = 2.52

We have, pH + pOH =14

pOH = 14 - pH = 14-2.52 = 11.48

We have, [H₃O ⁺ ] [OH ^-]= 10 ^-14

Therefore, [OH ^-]= 10 ^-14 / 0.003 = 3.33 10 ^-12 M

ANSWER : pH = 2.52 , pOH = 11.48 , [H₃O ⁺ ] =0.003 M , [OH ^-] = 3.33 10 ^-12 M