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The cis and Trans-forms of hyponitrous acid are given below in the attached document.
Hyponitrous acid has the chemical formula H2N2O2 or HON=NOH. 62.028 g/mol is the molecular weight of it. Additionally, it can take either a trans or cis form. When dry, the trans-hyponitrous acid crystallizes into white, explosive particles. Additionally, it has a half-life of 16 days at 25oC at a pH 1-3 in aqueous solution and is a weak acid (pKa1=7.21, pKa2=11.54). It decomposes into nitrous oxide and water. It is incorrect to think of N2O as the anhydride of H2N2O2. However, even though cis acid is unknown, we can still get its sodium salts.
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A 1.0-L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.
Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.
Answer:
The pH of this solution = 5.06
Explanation:
Given that:
number of moles of CH3COOH = 0.100 mol
volume of the buffer solution = 1.0 L
number of moles of NaC2H3O2 = 0.100 mol
The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.
we know that concentration in mole = Molarity/volume
Then concentration of [CH3COOH] = = 0.10 M
The chemical equation for this reaction is :
The conjugate base is CH3COO⁻
The concentration of the conjugate base [CH3COO⁻] is =
= 0.10 M
where the pka (acid dissociation constant)for CH3COOH = 4.74
If 0.035 mol of NaOH is added to the original buffer, the concentration of NaOH added will be = = 0.035 M
The ICE Table for the above reaction can be constructed as follows:
Initial 0.10 0.035 0.10 -
Change -0.035 -0.035 + 0.035 -
Equilibrium 0.065 0 0.135 -
By using Henderson-Hasselbalch equation:
The pH of this solution = pKa + log
The pH of this solution = 4.74 + log
The pH of this solution = 4.74 + log (2.076923077 )
The pH of this solution = 4.74 + 0.3174
The pH of this solution = 5.0574
The pH of this solution = 5.06 to two decimal places