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Chemistry · EQUILIBRIUM
हिन्दीAn acidic buffer contains equal molar concentrations of \( \text{HA} \) (\( pK_a = 4.40 \)) and its conjugate base \( \text{A}^- \). To change the pH of 1.0 L of this buffer to 5.00, the ratio of moles of strong base (\( \text{NaOH} \)) that must be added to the initial moles of \( \text{HA} \) present is: (Assume no volume change)
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Correct answer · Option A
0.60
Explanation
Initially, let \( n_{\text{HA}} = n_{\text{A}^-} = n_0 \). By Henderson equation, \( pH = pK_a + \log\frac{n_{\text{A}^-} + x}{n_{\text{HA}} - x} \implies 5.00 = 4.40 + \log\frac{n_0 + x}{n_0 - x} \). Thus \( \log\frac{n_0 + x}{n_0 - x} = 0.60 \implies \frac{n_0 + x}{n_0 - x} = 4 \). Solving gives \( n_0 + x = 4n_0 - 4x \implies 5x = 3n_0 \implies \frac{x}{n_0} = 0.60 \).
