Science, Tech, Math › Science Polyprotic Acid Example Chemistry Problem How to Work a Polyprotic Acid Problem Share Flipboard Email Print Sulfuric acid is an example of a common polyprotic acid. LAGUNA DESIGN / Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Anne Marie Helmenstine, Ph.D. Chemistry Expert Ph.D., Biomedical Sciences, University of Tennessee at Knoxville B.A., Physics and Mathematics, Hastings College Dr. Helmenstine holds a Ph.D. in biomedical sciences and is a science writer, educator, and consultant. She has taught science courses at the high school, college, and graduate levels. our editorial process Facebook Facebook Twitter Twitter Anne Marie Helmenstine, Ph.D. Updated February 05, 2019 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, it's 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, it's 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