Concept explainers
(a)
Interpretation:
The favourable orientation of the H2O molecule when approaching an anion has to be stated. The magnitude of the electric field experienced by the anion when the water dipole is 1.0 nm from the ion has to be calculated.
Concept introduction:
Polarized bonds are a result of the electronegativity difference between bonding atoms. The more electronegative atom acquires a partial negative charge and the less electronegative atom acquires a partial positive charge. This results in the formation of a dipole in the bond between the two atoms.
(a)
Answer to Problem 14B.4P
The magnitude of the electric field experienced by the anion when the water dipole is 1.0 nm from the ion is 1.11 ×108 V m−1_.
Explanation of Solution
The positive end of the dipole will lie closer to the negative anion when the H2O molecule approaches an anion.
The expression for the electric field (E) generated at a distance r is shown below.
E=μ2πε0r3 (1)
Where,
- ε0 is the vacuum permittivity. (8.854×10−12 J−1 C2 m−1)
- μ is the dipole moment.
The value of μ is 1.85 D.
The conversion of D to C m is done as shown below.
1 D=3.34×10−30 C m
Therefore, the conversion of 1.85 D to C m is done as shown below.
1.85 D=1.85 ×3.34×10−30 C m=6.179×10−30 C m
The value of r is 1.0 nm.
The conversion of nm to m is shown below.
1 nm=10−9 m
Substitute the value of ε0 , r and μ in equation (1).
E=6.179×10−30 C m2×3.14×8.854×10−12 J−1 C2 m−1×(10−9 m)3=6.179×10−30 C m5.5603×10−38 J−1 C2 m2 =1.11×108 J C−1 m−1 (1 J C−1=1 V)= 1.11×108 V m−1 _
Therefore, the magnitude of the electric field experienced by the anion is 1.11 ×108 V m−1_.
(b)
Interpretation:
The magnitude of the electric field experienced by the anion when the water dipole is 0.3 nm from the ion has to be calculated.
Concept introduction:
Same as concept introduction in part (a).
(b)
Answer to Problem 14B.4P
The magnitude of the electric field experienced by the anion when the water dipole is 0.3 nm from the ion is 4.12×109 V m−1_.
Explanation of Solution
The positive end of the dipole will lie closer to the negative anion when the H2O molecule approaches an anion.
The expression for the electric field (E) generated at a distance r is shown below.
E=μ2πε0r3 (1)
Where,
- ε0 is the vacuum permittivity. (8.854×10−12 J−1 C2 m−1)
- μ is the dipole moment.
The value of μ is 1.85 D.
The conversion of D to C m is done as shown below.
1 D=3.34×10−30 C m
Therefore, the conversion of 1.85 D to C m is done as shown below.
1.85 D=1.85 ×3.34×10−30 C m=6.179×10−30 C m
The value of r is 0.3 nm.
The conversion of nm to m is shown below.
1 nm=10−9 m
Therefore, the conversion of 0.3 nm to m is shown below.
0.3 nm=0.3×10−9 m
Substitute the value of ε0 , r and μ in equation (1).
E=6.179×10−30 C m2×3.14×8.854×10−12 J−1 C2 m−1×(0.3×10−9 m)3=6.179×10−30 C m1.5013×10−39 J−1 C2 m2 =4.12×109 J C−1 m−1 (1 J C−1=1 V)= 4.12×109 V m−1_
Therefore, the magnitude of the electric field experienced by the anion is 4.12×109 V m−1_.
(c)
Interpretation:
The magnitude of the electric field experienced by the anion when the water dipole is 30 nm from the ion has to be calculated.
Concept introduction:
Same as concept introduction in part (a).
(c)
Answer to Problem 14B.4P
The magnitude of the electric field experienced by the anion when the water dipole is 30 nm from the ion is 4.12×103 V m−1_.
Explanation of Solution
The positive end of the dipole will lie closer to the negative anion when the H2O molecule approaches an anion.
The expression for the electric field (E) generated at a distance r is shown below.
E=μ2πε0r3 (1)
Where,
- ε0 is the vacuum permittivity. (8.854×10−12 J−1 C2 m−1)
- μ is the dipole moment.
The value of μ is 1.85 D.
The conversion of D to C m is done as shown below.
1 D=3.34×10−30 C m
Therefore, the conversion of 1.85 D to C m is done as shown below.
1.85 D=1.85 ×3.34×10−30 C m=6.179×10−30 C m
The value of r is 30 nm.
The conversion of nm to m is shown below.
1 nm=10−9 m
Therefore, the conversion of 30 nm to m is shown below.
30 nm=30×10−9 m=3×10−8 m
Substitute the value of ε0 , r and μ in equation (1).
E=6.179×10−30 C m2×3.14×8.854×10−12 J−1 C2 m−1×(3×10−8 m)3=6.179×10−30 C m1.5013×10−33 J−1 C2 m2 =4.12×103 J C−1 m−1 (1 J C−1=1 V)= 4.12×103 V m−1_
Therefore, the magnitude of the electric field experienced by the anion is 4.12×103 V m−1_.
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Chapter 14 Solutions
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