Answer:
Ag₂CrO₄(s) + H⁺(aq) ⟶ 2Ag⁺(aq) + HCrO₄⁻(aq)
Explanation:
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq).
Silver chromate is the salt of a strong base (AgOH) and a weak acid (H₂CrO₄).
HCrO₄⁻ is an even weaker acid than H₂CrO₄, so CrO₄²⁻ is a strong base.
Any added H⁺ will immediately combine with the chromate ions according to the reaction
H⁺ + CrO₄²⁻ ⟶ HCrO₄⁻
thereby removing chromate ions from solution.
According to Le Châtelier's Principle, more silver chromate will dissolve to replace the chromate ions that the H⁺ removes.
The overall equation for the reaction is
Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + <em>CrO₄²⁻(aq)
</em>
<u>H⁺(aq) + </u><em><u>CrO₄²⁻(aq)</u></em><u> ⟶ HCrO₄⁻(aq)
</u>
Ag₂CrO₄(s) + H⁺(aq) ⟶ 2Ag⁺(aq) + HCrO₄⁻(aq)
Answer: 9.98
Explanation:
1) The equation for the dissociation of pyridine is:
C₅H₅N₅(aq )+ H₂O(l) ⇄ C₅H₅NH⁺(aq) + OH⁻(aq)
2) Kb equation:
Kb = [C₅H₅NH⁺(aq)] [OH⁻(aq)] / [C₅H₅N₅(aq )]
Where:
[C₅H₅NH⁺(aq)] = [OH⁻(aq)] ← from the equilibrium reaction
[C₅H₅N₅(aq )] = 4.8 M ← from the statement
⇒ 1.9 × 10 ⁻⁹ = x² / 4.8 ⇒ x² = 9.12 × 10⁻⁹
⇒ x = 9.55 × 10⁻⁵ = [OH⁻(aq)]
3) pOH
pOH = - log [OH⁻(aq)] = 4.02
4) pOH + pH = 14
⇒ pH = 14 - 4.02 = 9.98