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2 answers:
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Explanation:
Moles of phosphorus pentachloride present initially = 2.5 mol
Moles of phosphorus trichloride at equilibrium = 0.338 mol
Initially
2.5 mol 0 0
At equilibrium:
(2.5 - x) mol x x
So, from above, the moles of phosphorus trichloride at equilibrium , x= 0.338 mol
Mass of 0.338 moles of phosphorus trichloride at equilibrium:
= 0.338 mol × 137.5 g/mol = 46.475 g
Moles of phosphorus pentachloride present at equilibrium :
= (2.5 - 0.338) mol = 2.162 mol
Mass of 2.162 moles of phosphorus pentachloride at equilibrium:
= 2.162 mol × 208.5 g/mol = 450.777 g
Moles of chloride gas present at equilibrium : 0.338 mol
Mass of 0.338 moles of chloride gas at equilibrium:
= 0.338 mol × 71 g/mol = 23.998 g
Answer:
- <em>The solution that has the highest concentration of hydroxide ions is </em><u>d. pH = 12.59.</u>
Explanation:
You can solve this question using just some chemical facts:
- pH is a measure of acidity or alkalinity: the higher the pH the lower the acidity and the higher the alkalinity.
- The higher the concentration of hydroxide ions the lower the acidity or the higher the alkalinity of the solution, this is the higher the pH.
Hence, since you are asked to state the solution with the highest concentration of hydroxide ions, you just pick the highest pH. This is the option d, pH = 12.59.
These mathematical relations are used to find the exact concentrations of hydroxide ions:
- pH + pOH = 14 ⇒ pOH = 14 - pH
- pOH = - log [OH⁻] ⇒
Then, you can follow these calculations:
Solution pH pOH [OH⁻]
a. 3.21 14 - 3.21 = 10.79 antilogarithm of 10.79 = 1.6 × 10⁻¹¹
b. 7.00 14 - 7.00 = 7.00 antilogarithm of 7.00 = 10⁻⁷
c. 7.93 14 - 7.93 = 6.07 antilogarithm of 6.07 = 8.5 × 10⁻⁷
d. 12.59 14 - 12.59 = 1.41 antilogarithm of 1.41 = 0.039
e. 9.82 14 - 9.82 = 4.18 antilogarithm of 4.18 = 6.6 × 10⁻⁵
From which you see that the highest concentration of hydroxide ions is for pH = 12.59.
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of =
= 0.0458 M
Concentration of =
= 0.0521 M
GIven that :
Ka =
Thus; it's pKa = 4.72
pH ≅ 4.80