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
We will need to see what the picture looks like post the picture of the molecule
Equilibrium is achieved when the reaction rate of the forward and backward reaction are equal or the concentrations of the reactants and the products are in an unchanging ratio. Specifically, this system is in a dynamic equilibrium.
<span>6.03 moles.
1 molecule of butane contains 4 carbon atoms and ten hydrogen atoms.
The molar mass is 4 times the atomic mass of carbon, 12 g/mol, plus 10 times the atomic weight of hydrogen, 1 g/mol.
Molar mass = 4 * 12 g/mol + 10 * 1 g/mol = 58 g/mol.
This means that 1 mole of butane has a mass of 58 g.
To figure out how many moles are in a sample of butane, divide the mass of sample in grams by 58 grams
Number of moles in sample = 350 g / 58 g/mol = 6.03 moles.</span>