For the diprotic weak acid H2A, Ka1 = 3.2 Solution These are the immediate equat
ID: 629870 • Letter: F
Question
For the diprotic weak acid H2A, Ka1 = 3.2Explanation / Answer
These are the immediate equation you have: H2A(aq) + H2O(l) ---> H3O+(aq) + HA-(aq) Ka1 = 0.0010 HA-(aq) + H2O(l) ----> H3O+(aq) + A2-(aq) Ka2 = 0.000046 Ka1 = [H3O+][HA-]/[H2A] Ka2 = [H3O+][A2-]/[HA-] [H3O+] = 10^-1.785 = 0.0164 M What you do here is solve for [HA-] in the Ka2 equation and substitute it into the Ka1 equation: Ka1Ka2 = ([H3O+]^2)[A2-]/[H2A] You know that the equilibrium concentration of tartaric acid has to equal its initial concentration (C(H2A)) minus the concentrations of [HA-] and [A2-] or: C(H2A) = [H2A] + [HA-] + [A2-] Substitute the equilibrium equation for Ka2 into [HA-]: C(H2A) = [H2A] + [H3O+][A2-]/Ka2 + [A2-] Now, solve for [H2A]: [H2A] = C(H2A) - ([H3O+]/Ka2 + 1)[A2-] Finally, plug in this expression into Ka1Ka2 = ([H3O+]^2)[A2-]/[H2A] and solve for [A2-] to get: (Ka1Ka2) = ([H3O+]^2)[A2-]/{C(H2A) - ([H3O+]/Ka2 + 1)[A2-]} At this point you can plug-in all the values to get a more pleasing expression to look at: (4.6x10^-8) = (0.00027)[A2-]/{0.285 - 357.5[A2-]} Multiply both sides by {0.285 - 357.5[A2-]} and distribute to get: (1.3x10^-8) - (1.6x10^-5)[A2-] = (0.00027)[A2-] Bringing both like-terms to the same side yields: (1.3x10^-8) = (2.9x10^-4)[A2-] and finally solve for [A2-]: [A2-] = (1.3x10^-8)/(2.9x10^-4) = 4.5x10^-5 I hope you were able to follow all the math.