A deck of cards has dimensions 8.9cm x 5.72cm x 1.82 cm. What is the volume of the deck in cubic centimeters?
1 answer:
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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!
Answer:
11.979 MHz
Explanation:
Answer:
Check the image in the file attached below.
Explanation:
The objective of this question is to use the diagram provided to indicate the distance that corresponds to the bond length of the N2 molecule by placing X on the horizontal axis.
The bond length shows the distance of the bonds between the nuclei of the bonded atoms between the N2 molecule and O2 molecule. From the diagram, we can see that the bond length of O2 shows a greater tendency over the N2 molecule and it also have higher potential energy than the N2 molecule.
