Calculate the degree of ionization of dimethylamine in its 0,02 M solution. `K_b=5.4 xx 10^(-4)`. What percentage of dimethylamine is ionized if the solution is also 0.01 M in NaOH ?

Answer

To solve the problem of calculating the degree of ionization of dimethylamine in a 0.02 M solution and the percentage of ionization when 0.01 M NaOH is also present, we can follow these steps:

### Step 1: Write the ionization equation for dimethylamine
Dimethylamine (CH₃)₂NH can ionize in water as follows:
\[ (CH₃)₂NH + H_2O \rightleftharpoons (CH₃)₂NH_2^+ + OH^- \]

### Step 2: Set up the equilibrium expression
The base dissociation constant \(K_b\) for dimethylamine is given by:
\[ K_b = \frac{[(CH₃)₂NH_2^+][OH^-]}{[(CH₃)₂NH]} \]
Given \(K_b = 5.4 \times 10^{-4}\) and the initial concentration of dimethylamine is 0.02 M, we can set up the expression at equilibrium.

### Step 3: Define the changes in concentration
Let \(x\) be the amount of dimethylamine that ionizes. At equilibrium:
- The concentration of \((CH₃)₂NH\) will be \(0.02 - x\)
- The concentration of \((CH₃)₂NH_2^+\) and \(OH^-\) will both be \(x\)

### Step 4: Substitute into the equilibrium expression
Substituting into the \(K_b\) expression:
\[ 5.4 \times 10^{-4} = \frac{x \cdot x}{0.02 - x} \approx \frac{x^2}{0.02} \]
(We can approximate \(0.02 - x \approx 0.02\) because \(x\) will be small.)

### Step 5: Solve for \(x\)
Rearranging gives:
\[ x^2 = 5.4 \times 10^{-4} \times 0.02 \]
\[ x^2 = 1.08 \times 10^{-5} \]
\[ x = \sqrt{1.08 \times 10^{-5}} \approx 0.00329 \, \text{M} \]

### Step 6: Calculate the degree of ionization
The degree of ionization (\(\alpha\)) is given by:
\[ \alpha = \frac{x}{C_0} = \frac{0.00329}{0.02} \approx 0.1645 \]

### Step 7: Convert to percentage
To find the percentage of ionization:
\[ \text{Percentage of ionization} = \alpha \times 100 \approx 16.45\% \]

### Step 8: Consider the effect of NaOH
When 0.01 M NaOH is added, the concentration of \(OH^-\) increases. This will shift the equilibrium to the left due to the common ion effect.

### Step 9: Set up the new equilibrium expression
Now, the equilibrium expression becomes:
\[ K_b = \frac{(0.01 + x)(x)}{0.02 - x} \]
Assuming \(x\) is small compared to 0.01, we can approximate:
\[ K_b = \frac{0.01 \cdot x}{0.02} \]
Substituting \(K_b\):
\[ 5.4 \times 10^{-4} = \frac{0.01 \cdot x}{0.02} \]

### Step 10: Solve for \(x\) again
Rearranging gives:
\[ x = \frac{5.4 \times 10^{-4} \cdot 0.02}{0.01} = 1.08 \times 10^{-3} \, \text{M} \]

### Step 11: Calculate the new degree of ionization
Now calculate the new degree of ionization:
\[ \alpha = \frac{x}{C_0} = \frac{1.08 \times 10^{-3}}{0.02} = 0.054 \]

### Step 12: Convert to percentage
Finally, convert to percentage:
\[ \text{Percentage of ionization} = 0.054 \times 100 \approx 5.4\% \]

### Final Answer
The degree of ionization of dimethylamine in a 0.02 M solution is approximately 16.45%, and the percentage of dimethylamine ionized in a solution that is also 0.01 M in NaOH is approximately 5.4%.

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