Multiple Choice
What is the pH of the solution when 20.00 mL of 0.1563 M aniline hydrochloride (C6H5NH3+Cl-) is titrated with 15.00 mL of 0.1249 M NaOH? (Ka for aniline = 2.40 x 10^-5)
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17. Acid and Base Equilibrium
pH of Weak Acids
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What is the pH of a 0.010 M HNO₂ solution, given that the acid dissociation constant (Ka) for HNO₂ is 4.0 x 10⁻⁴?
A
2.00
B
3.00
C
3.35
D
4.00
Verified step by step guidance1
Start by writing the chemical equation for the dissociation of nitrous acid (HNO₂) in water: HNO₂ ⇌ H⁺ + NO₂⁻.
Use the expression for the acid dissociation constant (Ka) to set up the equilibrium expression: Ka = [H⁺][NO₂⁻]/[HNO₂].
Assume that the initial concentration of HNO₂ is 0.010 M and that the change in concentration due to dissociation is 'x'. Therefore, at equilibrium, [H⁺] = x, [NO₂⁻] = x, and [HNO₂] = 0.010 - x.
Substitute these equilibrium concentrations into the Ka expression: 4.0 x 10⁻⁴ = (x)(x)/(0.010 - x).
Solve the quadratic equation for 'x', which represents the concentration of H⁺ ions. Once 'x' is found, calculate the pH using the formula: pH = -log₁₀[H⁺].
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