What is the pH of aqueous 0.10 M HNO₂? K _a(HNO₂) = 4.5×10^{- 4}

Question

Quizplus · Accepted Answer

The given compound is HNO₂, which is a weak acid. Using the given acid dissociation constant (K _a) of 4.5×10^{- 4}, we can set up the equation for the dissociation of the acid: HNO₂ + H2O → H3O+ + NO₂- The equilibrium constant expression is: K _a = [H3O+][NO₂-]/[HNO₂] Let x be the concentration of [H3O+] and [NO₂-] at equilibrium, and assume that [H3O+] << [HNO₂] (which is generally true for weak acids). Then, we can also assume that [HNO₂] = 0.10 M - x. Substituting these assumptions into the equilibrium constant expression and solving for x using the quadratic formula gives: x = (4.5×10^{- 4})(0.10 - x) / (1 + 4.5×10^{- 4}) x = 1.47×10S1U1P2 Therefore, the pH of the solution is: pH = -log[H3O+] pH = -log(1.47×10S1U1P2) pH = 2.17 Therefore, the answer is C.

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