A student titrates an unknown monoprotic acid with a NaOH solution from a buret. After the addition of 12.35 mL of NaOH, the pH of the solution read 5.22. The equivalence point is reached at 24.70 mL of NaOH. What is the Ka of the acid?

A 30.0 mL analyte is made up of 0.155 g of an unknown monoprotic acid. The equivalence point is reached after 19.7mL of a 0.050 M NaOH solution is titrated into the analyte. Based on this finding, determine the molar mass of the unknown acid.

A student titrates 0.3883 grams of an unknown monoprotic acid to the equivalence point with 5.198 mL of 0.1365 M NaOH (aq). The molecular mass of the unknown acid is: