Answer:
1. The pH of 1.0 M trimethyl ammonium (pH = 1.01) is lower than the pH of 0.1 M phenol (5.00).
2. The difference in pH values is 4.95.
Explanation:
1. The pH of a compound can be found using the following equation:
![pH = -log([H_{3}O^{+}])](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH_%7B3%7DO%5E%7B%2B%7D%5D%29%20)
First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.
<u>Trimethyl ammonium</u>:
We can calculate [H₃O⁺] using the Ka as follows:
(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺
1.0 - x x x
![Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5B%28CH_%7B3%7D%29_%7B3%7DN%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5B%28CH_%7B3%7D%29_%7B3%7DNH%5E%7B%2B%7D%5D%7D)

By solving the above equation for x we have:
x = 0.097 = [H₃O⁺]
<u>Phenol</u>:
C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺
1.0 - x x x
![Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BC_%7B6%7DH_%7B5%7DO%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BC_%7B6%7DH_%7B5%7DOH%5D%7D)


Solving the above equation for x we have:
x = 9.96x10⁻⁶ = [H₃O⁺]
![pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%28%5BH_%7B3%7DO%5E%7B%2B%7D%5D%29%20%3D%20-log%289.99%20%5Ccdot%2010%5E%7B-6%7D%29%20%3D%205.00%20)
Hence, the pH of 1.0 M trimethyl ammonium is lower than the pH of 0.1 M phenol.
2. The difference in pH values for the two acids is:
Therefore, the difference in pH values is 4.95.
I hope it helps you!
Answer:
International System of Units (SI)
Explanation:
I hope it helps :)
Complete Question
The complete question is shown on the first uploaded image
Answer:
The equilibrium constant is 
Explanation:
From the question we are told that
The chemical reaction equation is

The voume of the misture is
The molar mass of
is a constant with value of 
The molar mass of
is a constant with value of 
The molar mass of
is a constant with value of 
Generally the number of moles is mathematically given as

For 


For 


For 


Generally the concentration of a compound is mathematicallyrepresented as

For 
![Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}](https://tex.z-dn.net/?f=Concentration%5BFe_2%20O_3%5D%20%3D%20%5Cfrac%7B0.222125%7D%7B5.4%7D)
For 
![Concentration[H_2] = \frac{1.815}{5.4}](https://tex.z-dn.net/?f=Concentration%5BH_2%5D%20%3D%20%5Cfrac%7B1.815%7D%7B5.4%7D)

For 
![Concentration [H_2O] = \frac{0.12}{5.4}](https://tex.z-dn.net/?f=Concentration%20%5BH_2O%5D%20%3D%20%5Cfrac%7B0.12%7D%7B5.4%7D)

The equilibrium constant is mathematically represented as
![K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cfrac%7B%5Bconcentration%20%5C%20of%20%5C%20product%5D%7D%7B%5Bconcentration%20%5C%20of%20%5C%20reactant%20%5D%7D)
Considering 
And 
At equilibrium the


The temperature of a reaction causes its rate of reaction to increase because the heat inputted into the solution excites the electrons that make up the solution, therefore making them move faster, colliding more often with other molecules of the solution. This increase in collision rates causes the rate of reaction to increase.