Answer:
STEP I
This is the balanced equation for the given reaction:-
[tex] 2KOH_{(aq)} + H_2SO_4{}_{(aq)} \rightarrow K_2SO_4{}_{(aq)} + 2H_2O_{(l)} [/tex]
STEP II
The compounds marked with (aq) are soluble ionic compounds. They must be
broken into their respective ions.
see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).
On breaking them into their respective ions :-
- 2KOH -> 2K+ + 2OH-
- H2SO4 -> 2H+ + (SO4)2-
- K2SO4 -> 2K+ + (SO4)2-
STEP III
Rewriting these in the form of equation
[tex] \underline{\pmb{2K^+} }+ 2OH^- + 2H^+ + \pmb{\underline{{SO_4{}^{2-}}} \: \rightarrow \: \underline{\pmb{2K^+}}} + \underline{\pmb{SO_4{}^{2-}}} + 2H_2O [/tex]
STEP IV
Canceling spectator ions, the ions that appear the same on either side of the equation
(note: in the above step the ions in bold have gotten canceled.)
[tex] \boxed{ \mathfrak{ \red{ 2OH^-{}_{(aq)} + 2H^+{(aq.)} \rightarrow H_2O{}_{(l)}}}} [/tex]
This is the net ionic equation.
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[tex] \\ [/tex]
[tex]\mathfrak{\underline{\green{ Why\: KOH \:has\: been\: taken\: as\: aqueous ?}}} [/tex]
- KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.
[Alkali metal hydroxides are the only halides soluble in water ]
