- Calculate the pH of each of the following solutions.
a. 0.100 M HONH2 (Kb= 1.1×10^-8)
b. 0.100 M HONH3Cl
c. pure H2O
d. a mixture containing 0.100 M HONH2 and 0.100 M HONH3Cl - The breaking of a covalent bond in such a way that each atom separates with one electron of the shared pair is called __
a) Homolytic fission
b) Heterolytic fission
c) Monolytic fission
d) Morpholytic fission
The Correct Answer and Explanation is :
1. pH Calculations
a. 0.100 M HONH2 (Kb = 1.1 x 10^-8)
HONH2 (Hydroxylamine) is a weak base. To calculate the pH, we need to first calculate the pOH using the Kb value.
The equation for the dissociation of HONH2 in water is:
[
HONH2 + H2O \rightleftharpoons HONH3^+ + OH^-
]
The Kb expression for this reaction is:
[
Kb = \frac{[HONH3^+][OH^-]}{[HONH2]}
]
Let the concentration of OH⁻ produced be ( x ), so the concentration of HONH3⁺ is also ( x ), and the concentration of HONH2 at equilibrium will be ( 0.100 – x ).
The equation becomes:
[
1.1 \times 10^{-8} = \frac{x^2}{0.100 – x}
]
For weak bases, we can approximate ( 0.100 – x \approx 0.100 ), so the equation simplifies to:
[
1.1 \times 10^{-8} = \frac{x^2}{0.100}
]
Solving for ( x ):
[
x^2 = 1.1 \times 10^{-8} \times 0.100 = 1.1 \times 10^{-9}
]
[
x = \sqrt{1.1 \times 10^{-9}} = 3.32 \times 10^{-5} \, \text{M}
]
Now, calculate the pOH:
[
\text{pOH} = -\log(3.32 \times 10^{-5}) = 4.78
]
Finally, calculate the pH:
[
\text{pH} = 14 – \text{pOH} = 14 – 4.78 = 9.22
]
So, the pH of the 0.100 M HONH2 solution is 9.22.
b. 0.100 M HONH3Cl
HONH3Cl is a salt formed from the reaction of a weak base (HONH2) and a strong acid (HCl). The pH of such a solution is determined by the hydrolysis of HONH3⁺.
HONH3⁺ reacts with water as follows:
[
HONH3^+ + H2O \rightleftharpoons HONH2 + H3O^+
]
The Kb for HONH2 is given as ( 1.1 \times 10^{-8} ), so the Ka for HONH3⁺ can be calculated using:
[
Ka = \frac{K_w}{Kb} = \frac{1.0 \times 10^{-14}}{1.1 \times 10^{-8}} = 9.09 \times 10^{-7}
]
Now, use the Ka expression:
[
Ka = \frac{[HONH2][H3O^+]}{[HONH3^+]}
]
Let the concentration of ( H3O^+ ) produced be ( x ), so the concentration of HONH2 is also ( x ), and the concentration of HONH3⁺ will be ( 0.100 – x ).
[
9.09 \times 10^{-7} = \frac{x^2}{0.100}
]
Solving for ( x ):
[
x^2 = 9.09 \times 10^{-7} \times 0.100 = 9.09 \times 10^{-8}
]
[
x = \sqrt{9.09 \times 10^{-8}} = 9.52 \times 10^{-4} \, \text{M}
]
Now, calculate the pH:
[
\text{pH} = -\log(9.52 \times 10^{-4}) = 3.02
]
So, the pH of the 0.100 M HONH3Cl solution is 3.02.
c. Pure H2O
Pure water has a neutral pH of 7.00 at 25°C, as the concentration of H₃O⁺ and OH⁻ is equal.
d. A mixture containing 0.100 M HONH2 and 0.100 M HONH3Cl
This is a buffer solution. The pH of a buffer solution is calculated using the Henderson-Hasselbalch equation:
[
\text{pH} = \text{pKa} + \log \left( \frac{[A^-]}{[HA]} \right)
]
In this case, ( [A^-] ) is the concentration of HONH2 and ( [HA] ) is the concentration of HONH3⁺. The pKa is related to Ka:
[
\text{pKa} = -\log(9.09 \times 10^{-7}) = 6.04
]
Using the Henderson-Hasselbalch equation:
[
\text{pH} = 6.04 + \log \left( \frac{0.100}{0.100} \right) = 6.04 + 0 = 6.04
]
So, the pH of the mixture is 6.04.
2. Breaking of a covalent bond
The correct answer is:
a) Homolytic fission
Explanation:
Covalent bonds form when two atoms share a pair of electrons. In homolytic fission, this bond breaks in such a way that each atom retains one electron from the shared pair. This results in the formation of two free radicals. A free radical is a species with an unpaired electron.
For example, consider the bond between two chlorine atoms (Cl–Cl). If this bond undergoes homolytic fission, each chlorine atom will take one of the shared electrons, resulting in two chlorine radicals:
[
Cl–Cl \rightarrow Cl^\cdot + Cl^\cdot
]
In heterolytic fission, on the other hand, the bond breaks unevenly, with one atom taking both electrons from the shared pair. This leads to the formation of a cation and an anion. For example:
[
Cl–Cl \rightarrow Cl^+ + Cl^-
]
Homolytic fission is common in reactions involving free radicals, such as in many organic reactions like combustion or polymerization. It plays a crucial role in the initiation steps of radical chain reactions. Heterolytic fission, conversely, is more common in ionic reactions where the bond breaks and the electrons are transferred to one atom, resulting in ion formation.
Therefore, the process where each atom gets one electron from the shared pair is called homolytic fission.