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!
Answer:
No one is correct. The correct expression is:
Keq = [H₂]² . [O₂]² / [H₂O]²
Explanation:
To build the Keq expression in a chemical equilibrium you must consider the molar concentrations of reactants / products, and they must be elevated to the stoichiometric coefficient.
The balance reaction is:
<u>2</u> H₂O (g) ⇄ <u>2</u> H₂ (g) + O₂ (g)
Keq = [H₂]² . [O₂] / [H₂O]²
In opposite side: <u>2</u> H₂ (g) + O₂ (g) ⇄ <u>2</u> H₂O (g)
Keq = [H₂O]² / [H₂]² . [O₂]
We will use this formlula: Mass in grams = Number of moles x Molecular mass of 1 mole.
Since, we know the avagadro number is 6.02 x 10²³, we only have two unknown values left which are the molecular mass of CH3OH and its mole.
Molecular Mass: C = 12, H= 1, O = 16, since we have C=12, H4 = 4, O = 16, we will add them up: 12 + 4 + 16 =32
We know that one mole of anything = 6.02 x 10²³.
So we will use this formula to find the mole of methanol: Number of moles = Number of molecules / Avagadro number
Number of moles of CH3OH = (9.79 x 10^24)/6.02 x 10²³) = 16.263 moles.
Now we know that the molecular mass = 32 and the mole is = 16.263.
Now we can find its mass by using this formula: <span>Mass in grams = Number of moles x Molecular mass of 1 mole.
</span>
Mass in grams = 16.263 x 32 = 520g
Answer:
Coefficient = 1.58
Exponent = - 5
Explanation:
pH = 2.95
Molar concentration = 0.0796M
Ka = [H+]^2 / [HA]
Ka = [H+]^2 / 0.0796
Therefore ;
[H+] = 10^-2.95
[H+] = 0.0011220 = 1.122 × 10^-3
Ka = [H+] / molar concentration
Ka = [1.122 × 10^-3]^2 / 0.0796
Ka = (1.258884 × 10^-6) / 0.0796
Ka = 15.815 × 10^-6
Ka = 1.58 × 10^-5
Coefficient = 1.58
Exponent = - 5
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