Answer:
a) The pH of the solution is 12.13.
b) The pH of the solution is 12.17.
Explanation:
Ionic product of water =
![K_w=[H^+][OH^-]](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5BOH%5E-%5D)
![1.01\times 10^-{14}=[H^+][OH^-]](https://tex.z-dn.net/?f=1.01%5Ctimes%2010%5E-%7B14%7D%3D%5BH%5E%2B%5D%5BOH%5E-%5D)
Taking negative logarithm on both sides:
![-\log[1.01\times 10^-{14}]=(-\log [H^+])+(-\log [OH^-])](https://tex.z-dn.net/?f=-%5Clog%5B1.01%5Ctimes%2010%5E-%7B14%7D%5D%3D%28-%5Clog%20%5BH%5E%2B%5D%29%2B%28-%5Clog%20%5BOH%5E-%5D%29)
The pH is the negative logarithm of hydrogen ion concentration in solution.
The pOH is the negative logarithm of hydroxide ion concentration in solution.

a)
of NaOH.
Concentration of hydroxide ions:

So, ![[OH^-]=1\times [NaOH]=1\times 1.39\times 10^{-2} M=1.39\times 10^{-2} M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D1%5Ctimes%20%5BNaOH%5D%3D1%5Ctimes%201.39%5Ctimes%2010%5E%7B-2%7D%20M%3D1.39%5Ctimes%2010%5E%7B-2%7D%20M)
![pOH=-\log[1.39\times 10^{-2} M]=1.86](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B1.39%5Ctimes%2010%5E%7B-2%7D%20M%5D%3D1.86)


pH=13.99-1.86=12.13
b)
of NaOH.
Concentration of hydroxide ions:

So, ![[OH^-]=3\times [Al(OH)_3]=3\times 0.0051 M=0.0153 M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D3%5Ctimes%20%5BAl%28OH%29_3%5D%3D3%5Ctimes%200.0051%20M%3D0.0153%20M)
![pOH=-\log[0.0153 M]=1.82](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5B0.0153%20M%5D%3D1.82)


pH=13.99-1.82=12.17
When forming an ion, this sodium atom will lose 1 electron, as it is the easiest way to form a full shell, as opposed to gaining 7 electrons.
Answer:
I believe it is a hope this helps
What you were given is the balanced chemical equation