At what molar concentration, the pH of nitrous acid should be 2.0 ? `(K_a=4.5xx10^(-4))`

Answer

To find the molar concentration of nitrous acid (HNO₂) at which the pH is 2.0, we can follow these steps:

### Step 1: Calculate the concentration of hydrogen ions \([H^+]\)
The pH is defined as:
\[ \text{pH} = -\log[H^+] \]
Given that the pH is 2.0, we can find \([H^+]\):
\[ [H^+] = 10^{-\text{pH}} = 10^{-2} = 0.01 \, \text{M} \]

### Step 2: Set up the expression for the dissociation of nitrous acid
Nitrous acid dissociates in water as follows:
\[ \text{HNO}_2 \rightleftharpoons \text{H}^+ + \text{NO}_2^- \]
Let the initial concentration of nitrous acid be \(C\) and the degree of dissociation be \(\alpha\). At equilibrium, we have:
\[ [H^+] = C\alpha \]
Since we found \([H^+] = 0.01 \, \text{M}\), we can write:
\[ C\alpha = 0.01 \]

### Step 3: Use the acid dissociation constant \(K_a\)
The expression for the acid dissociation constant \(K_a\) is given by:
\[ K_a = \frac{[H^+][NO_2^-]}{[HNO_2]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{C(1-\alpha)} = \frac{\alpha^2}{1-\alpha} \]
Given \(K_a = 4.5 \times 10^{-4}\), we can set up the equation:
\[ 4.5 \times 10^{-4} = \frac{\alpha^2}{1 - \alpha} \]

### Step 4: Assume \(\alpha\) is small
If we assume \(\alpha\) is small (which is reasonable for weak acids), we can simplify \(1 - \alpha \approx 1\):
\[ 4.5 \times 10^{-4} \approx \alpha^2 \]
Thus,
\[ \alpha \approx \sqrt{4.5 \times 10^{-4}} \approx 0.0212 \]

### Step 5: Substitute \(\alpha\) back to find \(C\)
From the earlier equation \(C\alpha = 0.01\):
\[ C(0.0212) = 0.01 \implies C = \frac{0.01}{0.0212} \approx 0.4717 \, \text{M} \]

### Conclusion
The molar concentration of nitrous acid required to achieve a pH of 2.0 is approximately \(0.4717 \, \text{M}\).

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