You need to use the Ka for the acetic acid and the equilibrium equation.
Ka = 1.85 * 10^ -5
Equilibrium reaction: CH3COOH (aq) ---> CH3COO(-) + H(+)
Ka = [CH3COO-][H+] / [CH3COOH]
Molar concentrations at equilibrium
CH3COOH CH3COO- H+
0.50 - x x x
Ka = x*x / (0.50 - x) = x^2 / (0.50 - x)
Given that Ka is << 1 => 0.50 >> x and 0.50 - x ≈ 0.50
=> Ka ≈ x^2 / 0.50
=> x^2 ≈ 0.50 * Ka = 0.50 * 1.85 * 10^ -5 = 0.925 * 10^ - 5 = 9.25 * 10 ^ - 6
=> x = √ [9.25 * 10^ -6] = 3.04 * 10^ -3 ≈ 0.0030
pH = - log [H+] = - log (x) = - log (0.0030) = 2.5
Answer: 2.5
Answer:
a. ![K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5B%20FeSCN%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%7D%7B%5BFe%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%5BSCN%5E-_%7B%28aq%29%7D%5D%7D)
b. ![K_p = \dfrac{[H_2]^4}{[H_2O]^4}](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cdfrac%7B%5BH_2%5D%5E4%7D%7B%5BH_2O%5D%5E4%7D)
Explanation:
Untuk semua jenis reaksi umum:

Konstanta kesetimbangan ![K_c = \dfrac{[C]^c [D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5BC%5D%5Ec%20%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Dari pertanyaan yang diberikan:
a. 
Konstanta kesetimbangan:
![K_c = \dfrac{[ FeSCN^{3+}_{(aq)}] }{[Fe^{3+}_{(aq)}] [SCN^-_{(aq)}]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5B%20FeSCN%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%7D%7B%5BFe%5E%7B3%2B%7D_%7B%28aq%29%7D%5D%20%5BSCN%5E-_%7B%28aq%29%7D%5D%7D)
b. 
Konstanta kesetimbangan untuk tekanan parsial 
![K_p = \dfrac{[H_2]^4}{[H_2O]^4}](https://tex.z-dn.net/?f=K_p%20%3D%20%5Cdfrac%7B%5BH_2%5D%5E4%7D%7B%5BH_2O%5D%5E4%7D)
Karena Fe3O4 (s) hadir sebagai padatan.
The mass of titanium is = 47,867 g/1mol
Applying the rule of avrogado
1mol _______ 6,023 × 10^(23) at
0,075mol ___ x
X . 1mol = 0,075mol . 6,023 . 10^(23)at
X = 0,075 . 6,023 . 10^(23) at
X = 4,51 . 10^(22) atoms
Hope this helps
So the solution inside doesn't splash
Answer:
See explanation
Explanation:
Hello there!
In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:
![Ksp=[Ag^+][Cl^-]](https://tex.z-dn.net/?f=Ksp%3D%5BAg%5E%2B%5D%5BCl%5E-%5D)
And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

I - 0 0
C - +x +x
E - x x
Which leads to the following modified equilibrium expression:

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.
Regards!