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:
[tex] pH = -log([H_{3}O^{+}]) [/tex]
First, we need to find [H₃O⁺] for trimethyl ammonium and for phenol.
Trimethyl ammonium:
We can calculate [H₃O⁺] using the Ka as follows:
(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺
1.0 - x x x
[tex]Ka = \frac{[(CH_{3})_{3}N][H_{3}O^{+}]}{[(CH_{3})_{3}NH^{+}]}[/tex]
[tex] 10^{-pKa} = \frac{x*x}{1.0 - x} [/tex]
[tex] 10^{-9.80}(1.0 - x) - x^{2} = 0 [/tex]
By solving the above equation for x we have:
x = 0.097 = [H₃O⁺]
[tex] pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01 [/tex]
Phenol:
C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺
1.0 - x x x
[tex]Ka = \frac{[C_{6}H_{5}O^{-}][H_{3}O^{+}]}{[C_{6}H_{5}OH]}[/tex]
[tex] 10^{-10} = \frac{x^{2}}{1.0 - x} [/tex]
[tex] 1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0 [/tex]
Solving the above equation for x we have:
x = 9.96x10⁻⁶ = [H₃O⁺]
[tex] pH = -log([H_{3}O^{+}]) = -log(9.99 \cdot 10^{-6}) = 5.00 [/tex]
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:
[tex] \Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95 [/tex]
Therefore, the difference in pH values is 4.95.
I hope it helps you!
