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rusak2 [61]
3 years ago
14

Which of the following has the greatest electronegativity difference between the bonded atoms? View Available Hint(s) Which of t

he following has the greatest electronegativity difference between the bonded atoms? A strong acid made of hydrogen and a halogen, such as HCl A group 1 alkali metal bonded to fluoride, such as LiF. Carbon bonded to a group 6A (16) nonmetal chalcogen, such as in CO A diatomic gas, such as nitrogen (N2).
Chemistry
1 answer:
Reptile [31]3 years ago
7 0

Answer: A group 1 alkali metal bonded to fluoride, such as LiF.

Explanation:

Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself. The size of an atom increases as we move down the group because a new shell is added and electron gets added up.

1. A strong acid made of hydrogen and a halogen, such as HCl : A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms. Electronegativity difference = electronegativity of chlorine - electronegativity of hydrogen = 3-2.1= 0.9

2. A group 1 alkali metal bonded to fluoride, such as LiF: Ionic bond is formed when there is complete transfer of electron from a highly electropositive metal to a highly electronegative non metal.

Electronegativity difference = electronegativity of fluorine - electronegativity of lithium= 4-1= 3

3. Carbon bonded to a group 6A (16) nonmetal chalcogen, such as in CO: A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.

Electronegativity difference = electronegativity of oxygen - electronegativity of carbon= 3.5-2.5= 1.0

4. A diatomic gas, such as nitrogen (N_2): Non-polar covalent bond is defined as the bond which is formed when there is no difference of electronegativities between the atoms.

Electronegativity difference = 0

Thus the greatest electronegativity difference between the bonded atoms is in LiF.

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100.00 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution.
wolverine [178]

Answer:

a. pH = 2.04

b. pH = 3.85

c. pH = 8.06

d. pH = 11.56

Explanation:

The nitrous acid is a weak acid (Ka = 5.6x10⁻⁴) that reacts with NaOH as follows:

HNO₂ + NaOH → NaNO₂(aq) + H₂O(l)

a. At the beginning there is just a solution of 0.12M HNO₂. As Ka is:

Ka = [H⁺] [NO₂⁻] / [HNO₂]

Where [H⁺] and [NO₂⁻] ions comes from the same equilibrium ([H⁺] = [NO₂⁻] = X):

5.6x10⁻⁴ = X² / 0.15M

8.4x10⁻⁵ = X²

X = [H⁺] = 9.165x10⁻³M

As pH = -log [H⁺]

<h3>pH = 2.04</h3><h3 />

b. At this point we have HNO₂ and NaNO₂ (The weak acid and the conjugate base), a buffer. The pH of a buffer is obtained using H-H equation:

pH = pKa + log [NaNO₂] / [HNO₂]

<em>Where pH is the pH of the buffer,</em>

<em>pKa is -log Ka = 3.25</em>

<em>And [NaNO₂] [HNO₂] could be taken as the moles of each compound.</em>

<em />

The initial moles of HNO₂ are:

0.100L * (0.15mol / L) = 0.015moles

The moles of base added are:

0.0800L * (0.15mol / L) = 0.012moles

The moles of base added = Moles of NaNO₂ produced = 0.012moles.

And the moles of HNO₂ that remains are:

0.015moles - 0.012moles = 0.003moles

Replacing in H-H equation:

pH = 3.25 + log [0.012moles] / [0.003moles]

<h3>pH = 3.85</h3><h3 />

c. At equivalence point all HNO2 reacts producing NaNO₂. The volume added of NaOH must be 100mL. That means the concentration of the NaNO₂ is:

0.15M / 2 = 0.075M

The NaNO₂ is in equilibrium with water as follows:

NaNO₂(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq) + Na⁺

The equilibrium constant, kb, is:

Kb = Kw/Ka = 1x10⁻¹⁴ / 5.6x10⁻⁴ = 1.79x10⁻¹¹ = [OH⁻] [HNO₂] / [NaNO₂]

<em>Where [OH⁻] = [HNO₂] = x</em>

<em>[NaNO₂] = 0.075M</em>

<em />

1.79x10⁻¹¹ = [X] [X] / [0.075M]

1.34x10⁻¹² = X²

X = 1.16x10⁻⁶M = [OH⁻]

pOH = -log [OH-] = 5.94

pH = 14-pOH

<h3>pH = 8.06</h3><h3 />

d. At this point, 5mL of NaOH are added in excess, the moles are:

5mL = 5x10⁻³L * (0.15mol / L) =7.5x10⁻⁴moles NaOH

In 100mL + 105mL = 205mL = 0.205L. [NaOH] = 7.5x10⁻⁴moles NaOH / 0.205L =

3.66x10⁻³M = [OH⁻]

pOH = 2.44

pH = 14 - pOH

<h3>pH = 11.56</h3>
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