Answer:
pH = 2.03
Explanation:
The pH can be calculated using the following equation:
![pH = -log [H_{3}O^{+}]](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20)
(1)
The concentration of H₃O⁺ is calculated using the dissociation constant of the next reaction:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺
1.00 M
![K_{a} = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}](https://tex.z-dn.net/?f=%20K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%20)
Solving the above equation for H₃O⁺, we have:
![[H_{3}O^{+}] = \frac{Ka*[CH_{3}COOH]}{[CH_{3}COO^{-}]}](https://tex.z-dn.net/?f=%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7BKa%2A%5BCH_%7B3%7DCOOH%5D%7D%7B%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%20)
(2)
The dissociation constant is equal to:
Now, by solving the equation of the solubility product for Herbigon, we can find [CH₃COO⁻]:
CH₃COOX ⇄ CH₃COO⁻ + X⁺
5.00x10⁻³ M
![K_{sp} = [CH_{3}COO^{-}][X^{+}]](https://tex.z-dn.net/?f=%20K_%7Bsp%7D%20%3D%20%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%5BX%5E%7B%2B%7D%5D%20)
![[CH_{3}COO^{-}] = \frac{K_{sp}}{[X^{+}]} = \frac{9.40 \cdot 10^{-6}}{5.00 \cdot 10^{-3}} = 1.88 \cdot 10^{-3} M](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%20%3D%20%5Cfrac%7BK_%7Bsp%7D%7D%7B%5BX%5E%7B%2B%7D%5D%7D%20%3D%20%5Cfrac%7B9.40%20%5Ccdot%2010%5E%7B-6%7D%7D%7B5.00%20%5Ccdot%2010%5E%7B-3%7D%7D%20%3D%201.88%20%5Ccdot%2010%5E%7B-3%7D%20M)
By entering the values of [CH₃COO⁻] and Ka, into equation (2) we can calculate [H₃O⁺]:
![[H_{3}O^{+}] = \frac{1.74 \cdot 10^{-5}*[1.00]}{[1.88 \cdot 10^{-3}]} = 9.26 \cdot 10^{-3} M](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B1.74%20%5Ccdot%2010%5E%7B-5%7D%2A%5B1.00%5D%7D%7B%5B1.88%20%5Ccdot%2010%5E%7B-3%7D%5D%7D%20%3D%209.26%20%5Ccdot%2010%5E%7B-3%7D%20M)
Hence, the pH is:
![pH = -log [H_{3}O^{+}] = -log [9.26 \cdot 10^{-3}] = 2.03](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%20%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3D%20-log%20%5B9.26%20%5Ccdot%2010%5E%7B-3%7D%5D%20%3D%202.03%20)
Therefore, the pH must be 2.03 to yield a solution in which the concentration of X⁺ is 5.00x10⁻³M.
I hope it helps you!
Atomic mass is the number that you get when the protons and neutrons in the nucleus are added together atomic number is the number of protons in the nucleus
Answer:
0.500 moles of CO2 has a volume of 11.2 L at STP (option B)
Explanation:
Step 1: Data given
Volume of a gas at STP = 11.2 L
STP: Pressure = 1 atm and temperature = 273 K
Step 2: Calculate volume
p*V= n*R*T
V = (n*R*T)/p
⇒with V = the volume of the gas = TO BE DETERMINED
⇒with n = the number of moles of the gas
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 273 K
⇒with p = the pressure of the gas = 1 atm
A
) 0.250 mole of NH3
V = (0.250 * 0.08206 * 273) / 1
V = 5.6 L
B
) 0.500 mole of CO2
V = (0.500 * 0.08206 * 273) / 1
V = 11.2 L
C
) 0.750 mole of NH3
V = (0.750 * 0.08206 * 273) / 1
V = 16.8 L
D) 1.00 mole of CO2
V = (1.00 * 0.08206* 273) / 1
V = 22.4 L
0.500 moles of CO2 has a volume of 11.2 L at STP (option B)
Answer:
Explanation:
glucose-1-phosphate → glucose-6-phosphate, ΔGo = -7.28 kJ/mol
fructose-6-phosphate → glucose-6-phosphate, ΔGo = -1.67 kJ/mol
subtracting the equation
glucose-1-phosphate - fructose-6-phosphate = 0 , ΔGo = -7.28 - ( -1.67 ) kJ / mol
glucose-1-phosphate = fructose-6-phosphate ΔGo = - 5.61 kJ / mol
- ΔGo = RT lnK
5.61 x 10³ = 8.31 x 298 x lnK
lnK = 2.265
K = 9.63 .
