Polyprotic Acid Example Chemistry Problem

Polyprotic Acid Example Chemistry Problem A polyprotic acid is an acid that can donate more than one hydrogen atom (proton) in an aqueous solution. To find the pH of this type of acid, its necessary to know the dissociation constants for each hydrogen atom. This is an example of how to work a polyprotic acid chemistry problem. Polyprotic Acid Chemistry Problem Determine the pH of a 0.10 M solution of H2SO4. Given: Ka2 1.3 x 10-2 Solution H2SO4 has two H (protons), so it is a diprotic acid that undergoes two sequential ionizations in water: First ionization: H2SO4(aq) → H(aq) HSO4-(aq) Second ionization: HSO4-(aq) ⇔ H(aq) SO42-(aq) Note that sulfuric acid is a strong acid, so its first dissociation approaches 100%. This is why the reaction is written using → rather than ⇔. The HSO4-(aq) in the second ionization is a weak acid, so the H is in equilibrium with its conjugate base. Ka2 [H][SO42-]/[HSO4-] Ka2 1.3 x 10-2 Ka2 (0.10 x)(x)/(0.10 - x) Since Ka2 is relatively large, its necessary to use the quadratic formula to solve for x: x2 0.11x - 0.0013 0 x 1.1 x 10-2 M The sum of the first and second ionizations gives the total [H] at equilibrium. 0.10 0.011 0.11 M pH -log[H] 0.96 Learn More Introduction to Polyprotic Acids Strength of Acids and Bases Concentration of Chemical Species First Ionization H2SO4(aq) H+(aq) HSO4-(aq) Initial 0.10 M 0.00 M 0.00 M Change -0.10 M +0.10 M +0.10 M Final 0.00 M 0.10 M 0.10 M Second Ionization HSO42-(aq) H+(aq) SO42-(aq) Initial 0.10 M 0.10 M 0.00 M Change -x M +x M +x M At Equilibrium (0.10 - x) M (0.10 + x) M x M