The Molecule of Sodium Formate along with Formal Charges (in blue) and lone pair electrons (in red) is attached below.
Sodium Formate is an ionic compound made up of a positive part (Sodium Ion) and a polyatomic anion (Formate).
Nomenclature:
In ionic compounds the positive part is named first. As sodium ion is the positive part hence, it is named first followed by the negative part i.e. formate.
Name of Formate:
Formate ion has been derived from formic acid ( the simplest carboxylic acid). When carboxylic acids looses the acidic proton of -COOH, they are converted into Carboxylate ions.
E.g.
HCOOH (formic acid) → HCOO⁻ (formate) + H⁺
H₃CCOOH (acetic acid) → H₃CCOO⁻ (acetate) + H⁺
Formal Charges:
Formal charges are calculated using following formula,
F.C = [# of Valence e⁻] - [e⁻ in lone pairs + 1/2 # of bonding electrons]
For Oxygen:
F.C = [6] - [6 + 2/2]
F.C = [6] - [6 + 1]
F.C = 6 - 7
F.C = -1
For Sodium:
F.C = [1] - [0 + 0/2]
F.C = [1] - [0]
F.C = 1 - 0
F.C = +1
Answer:
11·699
Explanation:
Given the concentration of hydroxide ion in the solution is 5 ×
M
Assuming the temperature at which it is asked to find the pH of the solution be 298 K
<h3>At 298 K the dissociation constant of water is

</h3><h3>∴ pH + pOH = 14 at 298 K</h3><h3>pOH of the solution = -log( concentration of hydroxide ion )</h3>
∴ pOH of the given solution = - log(5 ×
= -0·699 + 3 = 2·301
pH of the given solution = 14 - 2·301 = 11·699
∴ pH of the solution = 11·699
Answer:
Ca(OH)₂ (aq) + 2NH₄Cl (aq) → CaCl₂ (aq) + 2NH₃ (g) + 2H₂O (l)
Ca²⁺ + 2OH⁻ + 2NH₄⁺ + 2Cl⁻ → Ca⁺² + 2Cl⁻ + 2NH₃ (g) + 2H₂O (l)
2OH⁻ + 2NH₄⁺ → 2NH₃ (g) + 2H₂O (l)
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