In this example, we calculate the concentration of an unknown triprotic acid based on experimental titration data. Our problem reads, "The titration of 7.50 milliliters of an unknown triprotic acid requires 17.9 milliliters of 0.225 molar potassium hydroxide solution to reach the equivalence point. What is the concentration of the unknown acid sample?" First, we need a balanced chemical equation for the acid-base reaction. We don't know which triprotic acid we're titrating, but what's important is that a triprotic acid has 3 hydrogen atoms that can be removed. We can represent the triprotic acid as H3A, where A is a generic anion with a 3- charge. In an acid-base neutralization reaction, an acid and a base react to produce water and a salt. The salt will be K3A: we know potassium has a 1+ charge, and the anion A has a 3- charge, so we need 3 potassium for every 1 A 3- anion in the formula unit. Now let's balance. In an acid-base reaction, 1 H+ from the acid reacts with 1 hydroxide from the base to produce 1 water. In this reaction, we have 3 H+ ions from the acid, so we need 3 hydroxide, and we will produce 3 water molecules. This reaction should now be balanced. We want to calculate the molarity of the acid; the equation for molarity is moles of solute divided by liters of solution. To find the moles of H3A, we can use dimensional analysis to convert our volume of potassium hydroxide solution to moles of H3A. First, we can convert from milliliters to liters of KOH solution. Then we can use the molarity of potassium hydroxide to convert from liters of potassium hydroxide solution to moles of potassium hydroxide. Then we can convert from moles of potassium hydroxide to moles of H3A using the coefficients in the balanced chemical equation. We need 3 moles of potassium hydroxide for every 1 mole of our triprotic acid. Completing the calculation, we have 1.343 times 10 to the negative third moles of H3A. We can substitute this into our equation for molarity. Moles of H3A go in the numerator, and liters of H3A solution go in the denominator. We convert our 7.50 milliliters of H3A solution into liters before plugging it into this equation. Completing our molarity calculation, we should get 0.179 molar H3A.