Answer:
HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)
Explanation:
Complete question
Dissolved hydrofluoric acid reacts with dissolved sodium hydroxide to form water and aqueous sodium fluoride. What is the net ionic equation
Equilibrium equation between the undissociated acid and the dissociated ions
HF(aq)⇌H+(aq)+F−(aq)
Sodium hydroxide will dissociate aqueous solution to produce sodium cations, Na+, and hydroxide anions, OH−
NaOH(aq)→Na+(aq)+OH−(aq)
Hydroxide anions and the hydrogen cations will neutralize each other to produce water.
H+(aq)+OH−(aq)→H2O(l)
On combining both the equation, we get –
HF(aq)+Na+(aq)+OH−(aq)→Na+(aq)+F−(aq)+H2O(l)
The Final equation is
HF(aq)+NaOH(aq)→NaF(aq)+H2O(l)
1 ba+2 br——>1 babr2
u just have to make sure u have the same number of each type of atom on either side of the equation:)
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:
PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32
