Answer:
4.96E-8 moles of Cu(OH)2
Explanation:
Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.
Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.
![pH= -log[H]\\pH= -log (\frac{kw}{[OH]})](https://tex.z-dn.net/?f=pH%3D%20-log%5BH%5D%5C%5CpH%3D%20-log%20%28%5Cfrac%7Bkw%7D%7B%5BOH%5D%7D%29)
![8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}](https://tex.z-dn.net/?f=8.23%20%3D%20-%20log%28%5Cfrac%7BKw%7D%7B%5BOH%5D%7D%20%5C%5C10%5E%7B-8.23%7D%20%3D%20Kw%2F%5BOH%5D%5C%5C%20%5BOH%5D%20%3D%20Kw%2F10%5E%7B-8.23%7D)
![[OH]=1.69E-6](https://tex.z-dn.net/?f=%5BOH%5D%3D1.69E-6)
This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:
In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":
The expression for Kps is:
![Kps= [Cu^{2+}] [OH]^2](https://tex.z-dn.net/?f=Kps%3D%20%5BCu%5E%7B2%2B%7D%5D%20%5BOH%5D%5E2)
The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.
"s" is the soluble quantity of Cu(OH)2.
The solution for this third grade equation is 
Now, let us calculate the moles in 1 L:
First, we convert the depth of the water into meters. This is:
60 feet = 18.3 meters
Now, we compute the additional pressure exerted due to the water, which is given by:
Pressure = density * gravitational field strength * height
P = 1000 * 9.81 * 18.3
P = 179.5 kPa
The atmosphere pressure is 101.325 kPa
The pressure of the gas bubbles 60 feet under water will be:
179.5 + 101.325 = 280.825 kPa
The pressure at the surface of the water will be equal to the atmospheric pressure, 101.325 kPa.
Because of this decrease in external pressure as gas bubbles rise, they are seen to expand.
Radius of Xenon = 1.3Ă—10â’8 cm
Volume = 100 ml = 0.1 L
Pressure P = 1.2 atm = 121.59 Kpa
Temperature = 281 K
R = Gas Constant = 8.31 J mol^-1 K^-1
Now find the number of atoms
PV = nRT => n = PV / RT
n = (121.59 x 0.1) / (8.31 x 281) = / 2335.11 = 0.0052
Number of atoms in a mole is same as Avogadro constant A, which is 6.02 x
10^23 particles.
n = number of atoms= 0.0052
N = number of particles
Avogadro constant A = 6.02 x 10^23
n = N/A => N = n x A = 0.0052 x 6.02 x 106^23 = 3.13 x 10^20
Volume of Xe atom which would be a sphere = (4/3) x pi x r^3
Volume = = (4/3) x 3.14 x (1.3Ă—10â’8)^3 = 9.2 x 10^-24
Volume occupied by these particles = n x Volume = 3.13 x 10^20 x 9.2 x
10^-24 = 0.00288
Fraction of volume will be = 0.00288 / 0.1 = 0.0288
Answer:
Correct option is A)
[H
+
]=
KaC
=
1.8×10
−6
=1.34×10
−3
pH=−log[H
+
]
=2.88
Explanation:
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