To find the \( K_a \) value of the weak acid, we can use the relationship between the pH, the concentration of the weak acid, and the \( K_a \).
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Calculate the \([H^+]\) from pH: \[ [H^+] = 10^{-\text{pH}} = 10^{-4.17} \approx 6.76 \times 10^{-5} , \text{M} \]
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Set up the equilibrium expression: Let's denote the weak acid as HA. We know that it dissociates in water according to the following equation: \[ HA \rightleftharpoons H^+ + A^- \] For a solution of concentration \( [HA]_0 = 0.0107 , \text{M} \):
- The change in concentration when dissociation occurs can be represented as \( -x \) for \( HA \) and \( +x \) for both \( H^+ \) and \( A^- \).
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Establish the concentrations at equilibrium:
- At equilibrium:
- \([HA] \approx [HA]_0 - x \approx 0.0107 - 6.76 \times 10^{-5} \approx 0.0106 , \text{M}\) (since \( x \) is very small compared to \( [HA]_0 \))
- \([H^+] = [A^-] = x \approx 6.76 \times 10^{-5} , \text{M}\)
- At equilibrium:
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Write the expression for \( K_a \): \[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(6.76 \times 10^{-5})(6.76 \times 10^{-5})}{0.0106} \]
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Calculate \( K_a \): First calculate \((6.76 \times 10^{-5})^2\): \[ (6.76 \times 10^{-5})^2 \approx 4.57676 \times 10^{-9} \]
Now substitute it into the \( K_a \) formula: \[ K_a = \frac{4.57676 \times 10^{-9}}{0.0106} \approx 4.32 \times 10^{-7} \]
Thus, the \( K_a \) value for the weak acid is approximately: \[ \boxed{4.32 \times 10^{-7}} \]