Αποτελέσματα Αναζήτησης
B: Half-equivalence point. pH at B = 4.74 = pKa. Figure 3. Titration of 25 mL of 0.1M acetic acid by 0.1M NaOH. Several differences are readily apparent in the comparison of Figures 1 and 3. There are variations in the initial pH, the rate of pH change, and the pH at the equivalence point.
If the equivalence point occurs at 15 mL, then the half-equivalence point occurs at 7.5 mL. Notice the relatively flat slope of the curve around the half-equivalence point. At this point
From the titration curve, the point denoted as half the equivalence point, where [HA] = [A−], has a pH = 5.7, and the ionization constant is calculated as: pH = p K
Consider the titration of a weak acid, HA, with a strong base that gives the following pH curve. On the curve, indicate the points that correspond to the following: The stoichiometric (equivalence point) The region with maximum buffering. pH = pKa.
At the half-equivalence point, the amount of still undissociated acid (HA) and the amount of conjugate base that has already been formed (A-) are equal to each other.
Before you begin calculations for any titration in CH 223, do the following: Determine if solution to be titrated contains an acid or a base, and determine if it is strong or weak. Determine if the titrant is an acid or base and if it is weak or strong.
Standardize a sodium hydroxide solution. Generate a titration curve for the titration of a weak acid with a strong base. Determine the equivalence point of the titration. Calculate the initial concentration of the weak acid. Calculate Ka for the weak acid using titration information.
