) 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, C6H5OH? What is the difference in pH values for the two acids?

Answers

Answer 1

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!


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Answer:

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Hello,

In this case, since the carbon of the initial compound is present in the carbon dioxide product, we can compute the mass and moles of carbon in the compound:

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Answers

Answer:

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