Answer : The concentrations of hydroxide and hydronium ions in a solution with a pH of 10.2 are,
and
respectively.
Explanation : Given,
pH = 10.2
pH : It is defined as the negative logarithm of the hydrogen ion concentration.
First we have to calculate the hydrogen ion concentration 
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
Now put the value of pH in this formula, we get the hydrogen ion concentration.
![10.2=-\log [H^+]](https://tex.z-dn.net/?f=10.2%3D-%5Clog%20%5BH%5E%2B%5D)
![[H^+]=6.3\times 10^{-11}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D6.3%5Ctimes%2010%5E%7B-11%7D)
Now we have to calculate the pOH of the solution.

Now put the value of pH, we get the value of pOH.



Now we have to calculate the hydroxide ion concentration 
![pOH=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D)
Now put the value of pOH in this formula, we get the hydroxide ion concentration.
![3.8=-\log [OH^-]](https://tex.z-dn.net/?f=3.8%3D-%5Clog%20%5BOH%5E-%5D)
![[OH^-]=1.58\times 10^{-4}](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1.58%5Ctimes%2010%5E%7B-4%7D)
Therefore, the concentrations of hydroxide and hydronium ions in a solution with a pH of 10.2 are,
and
respectively.
Explanation:
First Reaction;
Ca + ZnCl2 --> CaCl2 + Zn
Oxidized Reactant: Ca. There is increase in oxidation number from 0 to +2
Reduced Reactant: Zn. There is decrease in oxidation number form +2 to 0
Second Reaction:
FeI2 + Mg --> Fe + MgI2
Oxidized Reactant: Mg. There is increase in oxidation number from 0 to +2
Reduced Reactant: Fe. There is decrease in oxidation number form +2 to 0
Third Reaction;
Mg + 2AgNO3 --> Mg(NO3)2 + Ag
Oxidized Reactant: Mg. There is increase in oxidation number from 0 to +2
Reduced Reactant: Ag. There is decrease in oxidation number form +1 to 0