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?
17. Acid and Base Equilibrium
pH of Weak Acids
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Find the pH of a solution that is 9.0×10⁻² M in H₂CO₃. Express your answer using two decimal places.
A
4.56
B
5.89
C
3.74
D
2.15
Verified step by step guidance1
Identify that H₂CO₃ (carbonic acid) is a weak acid and can dissociate in water to form H⁺ ions and HCO₃⁻ ions. The dissociation can be represented by the equation: H₂CO₃ ⇌ H⁺ + HCO₃⁻.
Use the expression for the acid dissociation constant (Ka) for H₂CO₃. The first dissociation constant (Ka1) is typically used for calculating pH: Ka1 = [H⁺][HCO₃⁻]/[H₂CO₃].
Assume that the initial concentration of H₂CO₃ is 9.0×10⁻² M and that the change in concentration due to dissociation is 'x'. Therefore, at equilibrium, [H⁺] = [HCO₃⁻] = x and [H₂CO₃] = 9.0×10⁻² - x.
Substitute these equilibrium concentrations into the Ka expression: Ka1 = (x)(x)/(9.0×10⁻² - x). Since H₂CO₃ is a weak acid, assume x is small compared to 9.0×10⁻², simplifying the expression to Ka1 ≈ x²/(9.0×10⁻²).
Solve for x, which represents the [H⁺] concentration, and then calculate the pH using the formula: pH = -log₁₀([H⁺]).
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