Oxidation of Nitrogen in N2O3
1 answer:
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The complete table is inserted.
A table is given,
Formulas used:
pH= -log(H⁺)
pOH= -log(OH⁻)
pH+ pOH=14
Calculations:
For A: (H⁺)=2×10⁻⁸M
Using the pH formula:
pH= -log(H⁺)=-log(2×10⁻⁸)=7.69
pOH=14 - 7.69=6.3
Calculating OH concentration,
pOH= -log(OH⁻)
6.3= -log(OH⁻)
(OH⁻)=5.011×10⁻⁷M
Hence, the nature of A is basic.
Similarily,
For B,
(OH⁻)=1×10⁻⁷
Using the pH formula:
pOH= -log(OH⁻)= -log(1×10⁻⁷)=7
pH=14-7=7
Calculating H concentration,
pH= -log(H⁺)
7= -log(H⁺)
(H⁺)=1×10⁻⁷M
Hence, the nature of B is neutral.
Similarily,
For C,
pH=12.3
Using the pH formula:
pOH=14-12.3=1.7
Calculating H concentration,
pH= -log(H⁺)
12.3= -log(H⁺)
(H⁺)=5.011×10⁻¹³M
Calculating OH concentration,
pOH= -log(OH⁻)
1.7= -log(OH⁻)
(OH⁻)=1.99×10⁻²M
Hence, the nature of C is Basic.
Similarily,
For D,
pOH=6.8
Using the pH formula:
pH=14-6.8=7.2
Calculating H concentration,
pH= -log(H⁺)
7.2= -log(H⁺)
(H⁺)=6.309×10⁻⁸M
Calculating OH concentration,
pOH= -log(OH⁻)
6.8= -log(OH⁻)
(OH⁻)=1.58×10⁻⁷M
Hence, the nature of D is basic.
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Answer:
to the right (products side)
Explanation:
The equilibrium constant K describes the ratio between the concentration of products and reactants at equilibrium. For a general reaction:
a A + b B → c C + d D
The equilibrium constant expression is:
A low value of K indicates that the concentration of products (C and D) is low in relation with the concentration of reactants (A and B).
Conversely, a high value of K indicated that the concentration of products is high compared with the concentration of reactants.
Since K = 6.4 × 10⁹ is a high value, the concentration of products is higher than the concentration of reactants at equilibrium. Thus, the position of the equilibrium is favored to the right.