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:
1. KCLO3------>KCL + 3/2O2(g)
2. 122.5g/mol
3. 0.2mol
4. 18.5g
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D, the rate increases as concentrations increase.
Typically, reaction rates decrease with time because reactant concentrations decrease as reactions are converted to products. Reaction rates generally increase when reactant concentrations are increased.