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
The substances which will exist in molecular form in solution are-
A. CH₃NH₂
D. CH₃OH
The methyl amine (CH₃NH₂) and methanol (CH₃OH) is organic compounds which are hardly become ionic in nature and prefers to form as molecular state in the solution phase.
On the other hand B. LiOH and C. NH₄OH are inorganic compound and highly ionic in nature and remain as ionic state in solution.
Those are-
LiOH (s) → Li⁺(s) + OH⁻ (s)
NH₄OH→ NH₄⁺ (s) + OH⁻ (s)
Answer:
See explanation
Explanation:
Let us see what happens when each solution is mixed;
a) AlCl3(aq) + K3PO4(aq) ------> 3KCl(aq) + AlPO4(s)
A precipitate is formed here
b) RbCO3(aq) + NaCl(aq) -------> This is an impossible reaction hence no solid precipitate is formed here
c) MnCl2(aq) + Na2CO3(aq) → 2NaCl(aq) + MnCO3(s)
A precipitate is formed.
d) K2S(aq) + 2NH4Cl(aq) ------> 2KCl(aq) + (NH4)2S(aq)
No solid precipitate is formed
e) CaCl2(aq) + (NH4)2CO3(aq) → CaCO3(s) + 2NH4Cl(aq)
A solid precipitate is formed
B because you can feel the current
