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)
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
2 answers
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
HOBR==> H^+ + OBr^-
Ksp for HOBr=
S^2
=>S=8*10^-3.5
Now Ksp for NaOBr=S^2=6.4*10^-6
1 answer