# How to determine pKa from a titration graph

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Titration is a procedure that determines the concentration of a liquid, typically an acid or base. During titration, the substance of known concentration is added to the one whose concentration is unknown until their numbers of reacting molecules are equal.

Since the pH changes throughout an acid-base titration, you can plot pH vs. amount of "known" added. From this titration graph, you can calculate the Ka, which is the dissociation constant for the acid. The pKa is the negative logarithm of the Ka and allows you to express the Ka more conveniently than the Ka, which is often very small.

• Titration is a procedure that determines the concentration of a liquid, typically an acid or base.
• Since the pH changes throughout an acid-base titration, you can plot pH vs. amount of "known" added.

Plot your titration data, using either graph paper or Excel. The pH should be the y-axis, and the titrant (the reactant of known concentration) should be on the x-axis. If you are plotting by hand, try to be as accurate as possible. You should connect the dots to get an S-curve rather than creating a best-fit line.

Find the equivalence point, the pH at which the number of reacting molecules of acid and base were exactly equivalent. You probably used a colour indicator during your titration and already have this information. The equivalence point is also the steepest part of your titration curve.

Assign the pKa as the pH value halfway to the equivalence point. You can mathematically justify this because at this half-equivalence point, the concentrations of protonated and deprotonated unknown are equivalent to each other. The equilibrium expression for the unknown would be Ka = [unknown-][H3O+]/[Hunknown], and since the Hunknown concentration is equal to unknown-, this can be rewritten as Ka = [H3O+]. Take the logarithm of both sides, and multiply both sides by -1, and the equation becomes pKa = pH.

• Find the equivalence point, the pH at which the number of reacting molecules of acid and base were exactly equivalent.
• You can mathematically justify this because at this half-equivalence point, the concentrations of protonated and deprotonated unknown are equivalent to each other.