What is the net ionic equation when NaOBr is dissolved in water and the Ksp if the Ksp of the dissocataion of HOBr is 6.4 * 10^-6
The molecular equation is
NaOBr + HOH ==> NaOH + HOBr
Separate into ions.
Na^+ + OBr^- + HOH => Na^+ + OH^- + HOBr.
Note:You should place an (aq) after each ion to show it is in aqueous solution.)
Now cancel those ions common to both sides; i.e., Na^+ to leave the net ionic equation as
OBr^-(aq) + HOH(l) ==>OH^-(aq) + HOBr(aq)
Let [OBr^-]=S
HOBR==> H^+ + OBr^-
Ksp for HOBr=
S^2
=>S=8*10^-3.5
Now Ksp for NaOBr=S^2=6.4*10^-6
To find the net ionic equation when NaOBr is dissolved in water, we first need to know the dissociation reaction of NaOBr.
The formula for NaOBr is Na+OBr-. When it dissolves in water, it dissociates into its ions.
The dissociation reaction is:
NaOBr(s) ⟶ Na+(aq) + OBr-(aq)
Next, we need to consider the dissociation of HOBr. The dissociation reaction of HOBr is as follows:
HOBr(aq) ⟶ H+(aq) + OBr-(aq)
To write the net ionic equation, we cancel out the spectator ions, which are the ions that appear on both sides of the equation and do not participate in the reaction.
In this case, the spectator ion is OBr-. Therefore, the net ionic equation is:
HOBr(aq) ⟶ H+(aq)
Now, to calculate the Ksp (solubility product constant) for the dissociation of HOBr, we need the balanced equation for the dissociation of HOBr, which is already given above.
The Ksp expression for the dissociation of HOBr is:
Ksp = [H+][OBr-]
Given that the Ksp is 6.4 × 10^-6, we can conclude that Ksp = [H+][OBr-] = 6.4 × 10^-6.
To determine the net ionic equation when NaOBr is dissolved in water, we first need to know the dissociation reaction of NaOBr in water. The compound NaOBr can be dissociated into sodium (Na+) ions and hypobromite (OBr-) ions. The balanced molecular equation for this dissociation can be written as:
NaOBr (s) -> Na+ (aq) + OBr- (aq)
However, since we are interested in the net ionic equation, we need to eliminate the spectator ions that do not participate in the actual reaction. In this case, the sodium ions (Na+) are spectator ions because they appear on both sides of the equation. Therefore, the net ionic equation can be written as:
OBr- (aq) -> OBr- (aq)
In the net ionic equation, the hypobromite ion (OBr-) remains as the only participating species.
Now, let's discuss the Ksp (solubility product constant) of the dissociation of HOBr. The Ksp expresses the equilibrium constant for the dissociation of a sparingly soluble compound in water. In this case, we are given that the Ksp of the dissociation of HOBr is 6.4 * 10^-6.
The dissociation of HOBr in water can be represented as follows:
HOBr (s) ⇌ H+ (aq) + OBr- (aq)
The Ksp expression for this dissociation can be written as:
Ksp = [H+] * [OBr-]
Since the Ksp value is given as 6.4 * 10^-6, you can use this information to calculate the concentrations of H+ and OBr- ions. However, it is important to note that the Ksp value of HOBr does not directly relate to the net ionic equation involving NaOBr, as they are different species with different dissociation reactions.