Non metals acid subcategories
Answer:
<u>STEP I</u>
This is the balanced equation for the given reaction:-

<u>STEP II</u>
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-
<u>STEP III</u>
Rewriting these in the form of equation

<u>STEP </u><u>IV</u>
Canceling spectator ions, the ions that appear the same on either side of the equation
<em>(note: in the above step the ions in bold have gotten canceled.)</em>

This is the net ionic equation.
____________________________

- 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 ]
<u>Answer:</u> The molar solubility of
is 
<u>Explanation:</u>
Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
The balanced equilibrium reaction for the ionization of calcium fluoride follows:

s 2s
The expression for solubility constant for this reaction will be:
![K_{sp}=[Pb^{2+}][I^-]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BPb%5E%7B2%2B%7D%5D%5BI%5E-%5D%5E2)
We are given:

Putting values in above equation, we get:

Hence, the molar solubility of
is 