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
CaCO3=100g
CaO=56g
56g of CaO=100g of CaCO3
1.12g of CaO= 100/56*1.12 g of CaCO3
=2g of CaCO3 (ans)
Answer:
First and seventeenth group.
Explanation:
Hydrogen is a special case as it has only one electron in its outermost orbital.
The hydrogen can lose or can accept electron easily.
Thus it can form positive ion similar to alkali metals and negative ion similar to halogens.
Thus it can fall into two groups
a) I group [Alkali metals]
b) 17th Group [Halogens]
