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Marizza181 [45]
3 years ago
9

) Do you think the pH of 1,0 M tri-methyl ammonium (CH3)3NH+, pKa = 9.80, will be higher or lower than that of 1.0 M phenol, C6H

5OH? What is the difference in pH values for the two acids?
Chemistry
1 answer:
Elanso [62]3 years ago
7 0

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^{+}])

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^{+}]}

10^{-pKa} = \frac{x*x}{1.0 - x}

10^{-9.80}(1.0 - x) - x^{2} = 0    

By solving the above equation for x we have:  

x = 0.097 = [H₃O⁺]

pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01                                      

<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]}

10^{-10} = \frac{x^{2}}{1.0 - x}

1.0 \cdot 10^{-10}(1.0 - x) - x^{2} = 0

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

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:

\Delta pH = pH_{C_{6}H_{5}OH} - pH_{(CH_{3})_{3}NH^{+}} = 5.00 - 1.01 = 4.95

Therefore, the difference in pH values is 4.95.

I hope it helps you!

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