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Multiple Choice
What is the pH of a 0.210 mol L⁻¹ solution of a weak monoprotic acid with a dissociation constant (Ka) of 3.9×10⁻²?
A
3.12
B
4.56
C
1.65
D
2.34
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1
Identify the given values: the concentration of the weak acid \( [HA] = 0.210 \text{ mol L}^{-1} \) and the acid dissociation constant \( K_a = 3.9 \times 10^{-2} \).
Write the expression for the acid dissociation constant: \( K_a = \frac{[H^+][A^-]}{[HA]} \).
Assume that the initial concentration of \( [H^+] \) and \( [A^-] \) is 0, and let \( x \) be the concentration of \( [H^+] \) and \( [A^-] \) at equilibrium. Therefore, \( [HA] \) at equilibrium is \( 0.210 - x \).
Substitute the equilibrium concentrations into the \( K_a \) expression: \( K_a = \frac{x^2}{0.210 - x} \).
Solve for \( x \) (which is \([H^+]\)) using the approximation method if \( x \) is small compared to 0.210, and then calculate the pH using \( \text{pH} = -\log[H^+] \).