Lead ii nitrate was added to a small test tube as well as Sodium hydroxide. What is the equilibrium equation for the reaction of lead ii ions and hydroxide ions?
The small ii s should be II.
and in parentheses as (II).
Pb^+2 + 2OH^- ==> Pb(OH)2(s).
If excess OH^- is added, the Pb(OH)2(s) dissolves to form a complex ion.
Pb(OH)2(s) + 2OH^- ==> Pb(OH)4-2
To determine the equilibrium equation for the reaction of lead II ions (Pb²⁺) and hydroxide ions (OH⁻), we can use the solubility rules and balanced chemical equations.
First, let's start by writing the chemical formulas for lead II nitrate and sodium hydroxide:
Lead II Nitrate: Pb(NO₃)₂
Sodium Hydroxide: NaOH
According to the solubility rules, lead II nitrate is soluble in water, and sodium hydroxide is also soluble in water. Therefore, both compounds will dissociate into their respective ions in an aqueous solution:
Pb(NO₃)₂ ⟶ Pb²⁺ + 2 NO₃⁻
NaOH ⟶ Na⁺ + OH⁻
Now, let's examine the reaction between the lead II ions and hydroxide ions to form a precipitate. The balanced equation for the reaction can be determined by combining the ions that have exchanged:
Pb²⁺ + 2 OH⁻ ⟶ Pb(OH)₂
So, the equilibrium equation for the reaction between lead II ions and hydroxide ions is:
Pb²⁺ + 2 OH⁻ ⟶ Pb(OH)₂