What is the pH of an aqueous 0.50 M KCN solution? K _a(HCN) = 6.2×10^{- 10}

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KCN is a salt that dissociates in water to form K+ and CN- ions. The CN- ion is the conjugate base of the weak acid HCN. The dissociation of HCN can be represented as HCN ⇌ H+ + CN-. The given Ka for HCN is 6.2×10^-10. To find the pH of the solution, we first need to calculate the Kb for CN- because it acts as a base in water. The relationship between Ka and Kb is given by the equation Ka × Kb = Kw, where Kw is the ion-product constant for water at 25°C, which is 1.0×10^-14.Kb for CN- = Kw / Ka = 1.0×10^-14 / 6.2×10^-10 = 1.61×10^-5Next, we use the formula for the base dissociation constant, Kb = [OH-][HB+] / [B], where [OH-] is the concentration of hydroxide ions, [HB+] is the concentration of the conjugate acid, and [B] is the concentration of the base. Assuming that the concentration of CN- is approximately equal to the initial concentration of KCN (0.50 M) because KCN is a strong electrolyte and fully dissociates, and that the change in concentration due to the dissociation of CN- is small, we can simplify the expression to solve for [OH-].Kb ≈ [OH-]^2 / [B]Rearranging and solving for [OH-], we get:[OH-] = sqrt(Kb × [B]) = sqrt(1.61×10^-5 × 0.50) = sqrt(8.05×10^-6) ≈ 2.84×10^-3 MTo find the pH, we first calculate the pOH:pOH = -log[OH-] = -log(2.84×10^-3) ≈ 2.55Finally, we use the relationship between pH and pOH, which is pH + pOH = 14, to find the pH:pH = 14 - pOH = 14 - 2.55 = 11.45

What Is the PH of an Aqueous 0