Calculate the pressure exerted by 66.0 g of CO2 gas at -14.5°C that occupies a volume of 50.0 L
1 answer:
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Answer:
The freezing point will be
Explanation:
The depression in freezing point is a colligative property.
It is related to molality as:
Where
Kf=
the molality is calculated as:
Depression in freezing point =
The new freezing point =
Answer:
* Before addition of any KOH:
pH = 0,0301
*After addition of 25.0 mL KOH:
pH = 1,30
*After addition of 50.0 mL KOH:
pH = 2,87
*After addition of 75.0 mL KOH:
pH = 6,70
*After addition of 100.0 mL KOH:
pH = 10,7
Explanation:
H₃PO₃ has the following equilibriums:
H₃PO₃ ⇄ H₂PO₃⁻ H⁺
k = [H₂PO₃⁻] [H⁺] / [H₃PO₃] k = 10^-(1,30) <em>(1)</em>
H₂PO₃⁻ ⇄ HPO₃²⁻ + H⁺
k = [HPO₃²⁻] [H⁺] / [H₂PO₃⁻] k = 10^-(6,70) <em>(2)</em>
Moles of H₃PO₃ are:
0,0500L×(1,8mol/L) = 0,09 moles of H₃PO₃
* Before addition of any KOH:
Using (1), moles in equilibrium are:
H₃PO₃: 0,09-x
H₂PO₃⁻: x
H⁺: x
Replacing:
4.51x10⁻³ - 0.050x -x² = 0
The right solution of x is:
x = 0.0466589
As volume is 0,050L
[H⁺] = 0.0466589moles / 0,050L = 0,933M
As pH = -log [H⁺]
<em>pH = 0,0301</em>
*After addition of 25.0 mL KOH:
0,025L×1,8M = 0,045 moles of KOH that reacts with H₃PO₃ thus:
KOH + H₃PO₃ → H₂PO₃⁻ + H₂O
That means moles of KOH will be the same of H₂PO₃⁻ and moles of H₃PO₃ are 0,09moles - 0,045moles = 0,045moles
Henderson-Hasselbalch formula is:
pH = pka + log₁₀ [A⁻] /[HA]
Where A⁻ is H₂PO₃⁻ and HA is H₃PO₃.
Replacing:
pH = 1,30 + log₁₀ [0,045mol] / [0,045mol]
<em>pH = 1,30</em>
*After addition of 50.0 mL KOH:
The addition of 50.0 mL KOH consume all H₃PO₃. Thus, in the solution you will have just H₂PO₃⁻. Thus, moles in solution for the equilibrium will be:
H₂PO₃⁻: 0,09-x
HPO₃²⁻: x
H⁺: x
Replacing:
1.8x10⁻⁸ - 2x10⁻⁷x - x² = 0
The right solution of x is:
x = 0.000134064
As volume is 50,0mL + 50,0mL = 100,0mL
[H⁺] = 0.000134064moles / 0,100L = 1.34x10⁻³M
As pH = -log [H⁺]
<em>pH = 2,87</em>
*After addition of 75.0 mL KOH:
Applying Henderson-Hasselbalch formula you will have 0,045 moles of both H₂PO₃⁻ HPO₃²⁻ and pka: 6,70:
pH = 6,70 + log₁₀ [0,045mol] / [0,045mol]
<em>pH = 6,70</em>
*After addition of 100.0 mL KOH:
You will have just 0,09moles of HPO₃²⁻, the equilibrium will be:
HPO₃²⁻ + H₂O ⇄ H₂PO₃⁻ + OH⁻ with kb = kw/ka = 1x10⁻¹⁴/10^-(6,70) = 5,01x10⁻⁸
kb = [H₂PO₃⁻] [OH⁻] / [HPO₃²⁻]
Moles are:
H₂PO₃⁻: x
OH⁻: x
HPO₃²⁻: 0,09-x
Replacing:
4.5x10⁻⁹ - 5.01x10⁻⁸x - x² = 0
The right solution of x is:
x = 0.000067057
As volume is 50,0mL + 100,0mL = 150,0mL
[OH⁻] = 0.000067057moles / 0,150L = 4.47x10⁻⁴M
As pH = 14-pOH; pOH = -log [OH⁻]
<em>pH = 10,7</em>
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I hope it helps!
Explanation:
(i) The equilibrium reaction equation will be as follows.
A reaction in which there will be absorption of heat energy is known as endothermic reaction.
A reaction in which there will be release of heat energy is known as exothermic reaction.
As liquid state of water is changing into solid state. So, it means that molecules of water came close to each other. Hence, there will be release of heat this means that reaction is exothermic in nature.
Hence, phase change from liquid to solid will be exothermic in nature.
Latent heat of fusion is defined as the amount of energy necessary to convert 1 gram of a solid into liquid state at its melting point.
So, when solid state of water changes into liquid state then it means energy is absorbed by the molecules of ice due to which they have gained kinetic energy. Hence, they moved away from each other leading to formation of liquid state of water.
Latent heat of freezing of liquid water is 334 J/g.
Specific heat of liquid water is 4.186
Specific heat of steam is 1.996
Specific heat of ice is 2.1
(ii) The equilibrium reaction equation will be as follows.
As liquid state of water is changing into gaseous or vapor state. So, it means molecules of liquid water has gained kinetic energy hence, they colloid more rapidly with each other.
As a result, heat will be absorbed by the liquid state of water. Hence, heat will be absorbed. Therefore, phase change from liquid to gas will be endothermic in nature.
Whereas when gaseous state of water will change into liquid state then heat will be released during this process of condensation. As a result, in that case reaction will be exothermic in nature.