A common type of chemistry experiment called titration determines the concentration of a substance dissolved in a solution. Acid-base titrations, in which an acid and a base neutralize each other, are the most common kind. The point at which all the acid or base in the analyte (the solution being analyzed) has been neutralized is called the equivalence point; depending on the acid or base in the analyte, some titrations will have a second equivalence point as well. You can calculate the pH of the solution at the second equivalence point easily.

- Pencil
- Paper
- Calculator
This calculation did not take the autoionization of water into account, which can become a factor in very dilute solutions of weak bases or acids. Nonetheless, it is a good estimate for these purposes and the kind of answer you'll be expected to give for this kind of problem.

Determine whether acid or base was present in the analyte, which kind of acid or base was present, and how much of it was present. If you're working on this question for a homework assignment, the information will be given to you. If, on the other hand, you have just done a titration in the lab, you will have collected the information as you performed the titration.

Remember that diprotic acids or bases (acids/bases that can donate or accept more than one hydrogen ion) are the kind that will have second equivalence points. Recall also that Ka1 is the equilibrium constant (ratio of products to reactants) for the first proton donation, while Ka2 is the equilibrium constant for the second proton donation. Look up the Ka2 for your acid or base in a reference text or online table (see Resources).

Determine the amount of conjugate acid or base in your analyte. This will be equivalent to the amount of acid or base originally present. Multiply the original analyte concentration by its volume. For example, suppose you start out with 40 mL of 1 molar oxalic acid. Convert the concentration to milliliters by dividing by 1000, then multiply this volume by its concentration. This will give you the number of moles of oxalic acid originally present: (40/1000) x 1= 0.04. There are 0.04 moles oxalic acid present.

Take the volume of titrant (the chemical you added during the titration) to neutralize the acid or base analyte and add it to the volume of analyte originally present. This will give you your final volume. For example, suppose that to reach second equivalence, 80 mL of 1 molar NaOH was added to 40 mL of 1 molar oxalic acid. The calculation will be 80 mL titrant + 40 mL analyte = 120 mL final volume.

Divide the number of moles of acid or base originally present in your analyte by the final volume. This will give you the final concentration of conjugate acid or base. For example, 120 mL was the final volume and 0.04 moles were originally present. Convert mL to liters and divide the number of moles by the number of liters: 120/1000 = 0.12 liters; 0.04 moles/0.12 liters = 0.333 moles per liter.

Determine the Kb of the conjugate base (or the Ka if it is a conjugate acid). Remember that the conjugate base is the species formed when you remove all the protons from an acid, while the conjugate acid is the species formed when you donate protons to a base. Consequently, at the 2nd equivalence point, the diprotic acid (oxalic acid, for example) will have been completely deprotonated and its Kb will be equal to 1 x 10^-14/the second Ka for oxalic acid. For a base, the Ka at the second equivalence point will be equal to 1 x 10^-14/the second Kb for the diprotic base. For example, oxalic acid was the analyte. Its Ka is 5.4 x 10^-5. Divide 1 x 10^-14 by 5.4 x 10^-5: (1 x 10^-14)/(5.4 x 10^-5) = 1.852 x 10^-10. This is the Kb for the completely deprotonated form of oxalic acid, the oxalate ion.

Set up an equilibrium constant equation in the following form: Kb = ([OH-][conjugate acid])/[conjugate base]. The square braces represent concentration.

Substitute x^2 for the two terms on top in the equation and solve for x as shown: Kb = x^2/[conjugate base]. For example, the concentration of sodium oxalate was 0.333 moles/L, and its Kb was 1.852 x 10^-10. When these values are plugged in, it yields the following calculation: 1.852 x 10^-10 = x^2/0.333. Multiply both sides of the equation by 0.333: 0.333 x (1.852 x 10^-10) = x^2; 6.167 x 10^-11 = x^2. Take the square root of both sides to solve for x: (6.167 x 10^-11)^1/2 = x. This yields the following: x = 7.85 x 10^-6. This is the concentration of hydroxide ions in the solution.

Convert from concentration of hydroxide ion or hydrogen ion to pH. If you have concentration of hydrogen ion, you just take the negative log to convert to pH. If you have concentration of hydroxide ion, take the negative log then subtract your answer from 14 to find the pH. For example, the concentration found was 7.85 x 10^-6 moles per liter of hydroxide ions: log 7.85 x 10^-6 = -5.105, therefore, -log 7.85 x 10^-6 = 5.105.

Subtract your answer from 14. For example, 14 - 5.105 = 8.90. The pH at the second equivalence point is 8.90.

#### Things You'll Need

#### Tips

References

- ChemBuddy: pH of Amphiprotic Salt
- "Chemical Principles: The Quest for Insight"; Peter Atkins, et al.; 2008
- Frostburg State University: What Is the pH at Equivalence Point?

Tips

- This calculation did not take the autoionization of water into account, which can become a factor in very dilute solutions of weak bases or acids. Nonetheless, it is a good estimate for these purposes and the kind of answer you'll be expected to give for this kind of problem.

About the Author

Based in San Diego, John Brennan has been writing about science and the environment since 2006. His articles have appeared in "Plenty," "San Diego Reader," "Santa Barbara Independent" and "East Bay Monthly." Brennan holds a Bachelor of Science in biology from the University of California, San Diego.

Photo Credits

Jupiterimages/Goodshoot/Getty Images