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If enough of a monoprotic acid is dissolved in water to produce a 0.0193 M solution with a pH of 6.59, what is the equilibrium constant, K, for the acid? 3.423 x10-12 K= Income
For weak acids that are so dilute or so weak that the pH of the solution lies between 6 and 7, the autoprotolysis of water must be taken into account when determining the K, value. At equilibrium there are four species in solution: the weak acid (HA), its conjugate (A-), H30+, and OH-. Start by developing four equations that describe what is occurring in the solution at equilibrium. First, write the Ky expression. Kw = [H3O+] [OH-] Then, write the K, expression for the weak acid (HA). Ka [H30+1 [A] [HA] All solutions must remain
< Feedback х All solutions must remain electroneutral. The charge balance is a way to describe the electroneutrality of the solution. The sum of the positive charges in solution equals the sum of the negative charges in solution. The charge balance for the solution is [H3O+] = [OH-] + [A-] = The material (mass) balance takes into account that the quantity of all species in solution containing an atom or group of atoms must equal the amount of that atom or group of atoms added to the solution initially. The material (mass) balance for the solution is [HA]initial = [HA] + [A-] You now have four equations and
You now have four equations and four unknowns. The value of Ky is known, as is the initial concentration of HA, 0.0193 M, which is given in the question. The concentration of [H3O+] can be determined from the pH. [H3O+] = 10-pH Therefore, the four unknowns are [HA], [AC], [OH-], and Kg. Rearrange the equations as outlined to develop an expression for K, in terms of the known variables, Kw, [HAlinitial, and [H3O+] 1. Rearrange the Ky expression to solve for [OH-] in terms of Kw and [H3071 2. Rearrange the charge balance to solve for IA) in terms of TO LOT