The pK_a of HCl is ?7. What is the value of the equilibrium constant, K_eq, for the following equilibrium? [The concentration of water in a dilute aqueous solution is 55 M] $\mathrm{HCl}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{\oplus}+\mathrm{Cl}^{\ominus}$ A) 1.8 × 10⁵ B) 385 C) 1.8 × 10?⁵ D) ?1.8 × 10⁵

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Quizplus · Accepted Answer

The equilibrium constant, K_eq, for the given equilibrium is:$$K_{eq}=\frac{[\mathrm{H}_{3} \mathrm{O}^{\oplus}][\mathrm{Cl}^{\ominus}]}{[\mathrm{HCl}]}$$The pK_a of HCl is -7, so the K_a of HCl is 10^-7. The K_a of HCl is related to the K_eq for the given equilibrium by the following equation:$$K_{a}=\frac{[\mathrm{H}_{3} \mathrm{O}^{\oplus}][\mathrm{Cl}^{\ominus}]}{[\mathrm{HCl}]} imes \frac{1}{[\mathrm{H}_{2} \mathrm{O}]}$$Substituting the given values into this equation, we get:$$10^{-7}=\frac{[\mathrm{H}_{3} \mathrm{O}^{\oplus}][\mathrm{Cl}^{\ominus}]}{[\mathrm{HCl}]} imes \frac{1}{55 \mathrm{M}}$$Solving for K_eq, we get:$$K_{eq}=\frac{[\mathrm{H}_{3} \mathrm{O}^{\oplus}][\mathrm{Cl}^{\ominus}]}{[\mathrm{HCl}]}=10^{-7} imes 55 \mathrm{M}=1.8 imes 10^{-5}$$Therefore, the value of K_eq for the given equilibrium is 1.8 × 10^-5.

The PKa of HCl Is ?7 HCl+H2O⇌H3O⊕+Cl⊖\mathrm{HCl}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{\oplus}+\mathrm{Cl}^{\ominus}HCl+H2​O⇌H3​O⊕+Cl⊖

The PK_a of HCl Is ?7 $\mathrm{HCl}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{\oplus}+\mathrm{Cl}^{\ominus}$