Answer:
HCOOH(aq) + OH-(aq) —> HCOO-(aq) + H2O(l)
Explanation:
HCOOH is a weak acid and so will not ionised completely in solution.
KOH is a strong base and will ionised completely as shown below
KOH(aq) –> K+(aq) + OH-(aq)
The overall reaction can be written as follow:
HCOOH(aq) + K+(aq) + OH-(aq) —> HCOO-(aq) + K+(aq) + H2O(l)
Cancel out the K+ to obtain the net ionic equation as shown below
HCOOH(aq) + OH-(aq) —> HCOO-(aq) + H2O(l)
The first shell is full at 2 electrons, the next 2 shells are full at 8 electrons, and any shells after that are full at 16 electrons
