Question

) 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?

Answer 1

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.

Step-by-step 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.

Trimethyl ammonium:

We can calculate [H₃O⁺] using the Ka as follows:

(CH₃)₃NH⁺ + H₂O → (CH₃)₃N + H₃O⁺

1.0 - x x x

`Ka = ([(CH_(3))_(3)N][H_(3)O^(+)])/([(CH_(3))_(3)NH^(+)])`

`10^(-pKa) = (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`

Phenol:

C₆H₅OH + H₂O → C₆H₅O⁻ + H₃O⁺

1.0 - x x x

`Ka = ([C_(6)H_(5)O^(-)][H_(3)O^(+)])/([C_(6)H_(5)OH])`

`10^(-10) = (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!