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⁺]
![pH = -log([H_{3}O^{+}]) = -log(0.097) = 1.01](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%280.097%29%20%3D%201.01%20)
<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!
Thomson used a beam of negatively charged particles. Using a beam of particles and detecting the scattering of the particles after they hit gold foil.
energy is required to move from one state or phase of matter to the next. Energy is used to make a liquid into a gas or a solid into a liquid.
Answer:
10 m
Explanation:
The mole fraction of FeCl₃ of 0.15, that is, per mole of solution, there are 0.15 moles of FeCl₃ and 1 - 0.15 = 0.85 moles of water.
The molar mass of water is 18.02 g/mol. The mass corresponding to 0.85 moles is:
0.85 mol × 18.02 g/mol = 15 g = 0.015 kg
The molality of FeCl₃ is:
m = moles of solute / kilogram of solvent
m = 0.15 mol / 0.015 kg
m = 10 m
Answer:
Procedure (2)
Explanation:
Assume the dialyses come to equilibrium in the allotted times.
Procedure (1)
If you are dialyzing 5 mL of sample against 4 L of water, the concentration of NaCl will be decreased by a factor of
Procedure (2)
For the first dialysis, the factor is
After a second dialysis, the original concentration of NaCl will be reduced by a factor of
Procedure (2) is more efficient by a factor of
