A solution is prepared that is initially 0.38M in ethylamine (C₂H5NH₂), a weak base, and 0.12M in ethylammonium chloride (C₂H5NH₂C1). Complete the reaction table below, so that you could use it to calculate the pH of this solution. Use x to stand for the unknown change in [OH-]. You can leave out the M symbol for molarity. initial change final [C,HẠNH,] [c,HẠNH; 0 0 0 0 [OH-] 0 0 0 Olo X S
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![A solution is prepared that is initially 0.38 M in ethylamine (\( \text{C}_2\text{H}_5\text{NH}_2 \)), a weak base, and 0.12 M in ethylammonium chloride (\( \text{C}_2\text{H}_5\text{NH}_3\text{Cl} \)). Complete the reaction table below, so that you could use it to calculate the pH of this solution.
Use \( x \) to stand for the unknown change in \([ \text{OH}^- ]\). You can leave out the \( M \) symbol for molarity.
| | \([ \text{C}_2\text{H}_5\text{NH}_2 ]\) | \([ \text{C}_2\text{H}_5\text{NH}_3^+ ]\) | \([ \text{OH}^- ]\) |
|------------------------|-------------------------------------------|-------------------------------------------|--------------------|
| **initial** | | | |
| **change** | | | |
| **final** | | | |
This table should be filled by using the initial concentrations and the changes during the reaction involving the species \([ \text{OH}^- ]\) represented by \( x \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa60cc6ce-11de-444a-af29-e9eb1b50e640%2F2b2b30de-b840-4a18-8c48-2d51fdaae4b3%2F3wsmroq_processed.png&w=3840&q=75)
In the question itself, the initial concentration of ethylamine and ethylammonium chloride is given.
It is given- [C2H5NH3Cl]= 0.12M
we know C2H5NH3Cl dissociates into C2H5NH3+ and Cl- . so we can write-
Initial concentration of [C2H5NH3+]=0.12M
Initial concentration of [C2H5NH2] = 0.38M
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