What is \( K_{\mathrm{a}} \) for \( \mathrm{HNO}_{2}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NO}_{2}(a q) \) ? A. \( \mathrm{K}_{\mathrm{a}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{NO}_{2^{-}}\right] \) B. \( \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{NO}_{2^{-}}\right]}{\left[\mathrm{NHO}_{2}\right]} \) C. \( \mathrm{K}_{\mathrm{a}}=\left[\mathrm{HNO}_{2}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{NO}_{2^{-}}\right] \) D. \( \mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{HNO}_{2}\right]}{\left[\mathrm{H}^{+}\right]\left[\mathrm{NO}_{2^{-}}\right]} \)
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The ionization of a weak acid in water is described by the equilibrium expression for its acid dissociation constant \( K_a \). In this case, the weak acid \( \mathrm{HNO}_{2} \) dissociates into \( \mathrm{H}^{+} \) and \( \mathrm{NO}_{2^{-}} \). Therefore, the correct expression for \( K_a \) accounts for the concentrations of products over the concentration of the reactant. The correct answer is: B. \( K_a = \frac{[\mathrm{H}^{+}][\mathrm{NO}_{2^{-}}]}{[\mathrm{HNO}_{2}]} \) This shows that the acid dissociation constant is a measure of how well the acid ionizes in solution!
