Experiment 4-008b: Molar Mass Determination via Acid-Base Titration
Report Sheets for Experiment 4-008b: pdf format
This
experiment has 3 parts:
Part
a) Preparation of a standard acid (KHP) solution
To qualify as a primary standard, a reagent must be stable and pure. The acid that will be used is potassium hydrogen phthalate a solid, monoprotic acid with the molecular formula HC8H4O4K (MW= 204.23 g/mole). Potassium hydrogen phthalate is generally abbreviate as “KHP”.
When dissolved in water, KHP dissociates into potassium cations and the hydrogen phthalate anion:
KHP(s) ----> K+(aq) + HC8H4O4-(aq)
The hydrogen phthalate anion is a weak acid, and when neutralized with OH, it undergoes an acid-base reaction:
HC8H4O4-(aq) + OH-(aq)---> C8H4O42-(aq) + H2O(l)
Given its purity and stability, it is a straightforward task to prepare a KHP solution of known concentration.
Part
b) Use of the KHP solution to standardize a solution of NaOH
After the KHP solution has been prepared, it will be used to standardize a solution of NaOH. The technique we will use is an acid-base titration.
After the NaOH solution has been standardized, inform the laboratory instructor what the concentration was determined to be. If the analysis is within acceptable limits ( 0.5% ) of the known value, proceed with the third part of the experiment. If not, let the laboratory instructor help improve your titration technique.
Part
c) Use of the standardized NaOH(aq)
to determine the molecular mass of an unknown acid.
Once the titration technique has been mastered, the standardized NaOH solution will be used to determine the concentration of a dilute solution of an unknown acid. When a known volume of standardized base is used in a titration, one can determine the number of moles of OH added. Since Molarity = (moles/liter), then moles = (Molarity)(Liter); both the molarity of the NaOH solution and the volume of solution (the “liters”) are known, so calculation of the moles of OH involved in a titration is not difficult.
For a monoprotic acid (such as HX), the moles of OH added during a titration equals the number of moles of acid initially present. Recognizing that moles = g/MM (where “g” = the grams of sample and “MM” = the molecular mass of the sample), it is not a giant leap to solve for “MM”
MM = g/moles
By controlling the number of grams of acid in the titration reaction, and by measuring the volume of standardized NaOH involved in the reaction (and thus the moles of acid involved), determination of the acid's molecular mass (MM) is straightforward.