Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. K pKa1 K pKa2 1.30
1 answer:
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!
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The given question is incomplete. The complete question is :
For the reaction at equilibrium, which statement correctly describes the effects of increasing pressure and adding
, respectively
a) Increasing pressure causes shift to reactants, adding causes shift to products.
b) Increasing pressure causes shift to products ,adding causes shift to reactants.
c) Increasing pressure causes shift to products, adding causes shift to products.
d) Increasing pressure causes shift to reactants,adding causes shift to reactants
Answer: Increasing pressure causes shift to reactants, adding causes shift to products.
Explanation:
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
For the given equation:
a) If the pressure is increased, the volume will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in pressure is taking place. As the number of moles of gas molecules is lesser at the reactant side. So, the equilibrium will shift in the left direction. i.e. towards reactants.
b) If is added, the equilibrium will shift in the direction where
is decreasing. So, the equilibrium will shift in the right direction. i.e. towards products.
Answer:
The concentration is 50,8 % w/v and radio strengths = 1,96.
Explanation:
Phenobarbital sodium is a medication that could treat insomnia, for example.
2,0 M of Phenobarbital sodium means 2 moles in 1L.
The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means 1g in <em>r</em> mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:
2 moles × = 508 g of Phenobarbital sodium.
1 L × = 1000 mL of solution
Thus, % w/v is:
× 100 = 50,8 % w/v
And radio strengths:
= 1,96. Thus, you have 1 g in 1,96 mL
I hope it helps!
Answer: 4.0 g/mL
Explanation:
The volume increased by 5.0 mL. Recall that the number of significant figures is equal to the number of certain values you can read plus one. Here, the volume increased from 14.0 mL to 19.0 mL, so the volume of X is 5.0 mL.
