Question

Consider the titration of a 23.0 ml sample of 0.105 moll-1 ch3cooh (ka=1.8×10-5) with 0.130 moll-1 naoh. Determine the pH at one-half of the equivalence point.

Answer 1

The pH at one-half of the equivalence point in the titration of CH3COOH with NaOH can be calculated using the Henderson-Hasselbalch equation. At one-half of the equivalence point, the concentration of CH3COOH is equal to the concentration of CH3COO-. Using the dissociation constant, Ka, for CH3COOH, the pH can be calculated to be 4.74.

Step-by-step explanation:

The pH at one-half of the equivalence point can be determined by considering the titration curve for the reaction between CH3COOH and NaOH. At one-half of the equivalence point, the number of moles of CH3COOH that have reacted is half of the initial number of moles. In this case, only 12.50 mL of NaOH has been added, so only half of the initial 0.00250 mol of CH3COOH has reacted.

To calculate the pH at one-half of the equivalence point, we need to look at the dissociation of CH3COOH. The dissociation constant, Ka, for CH3COOH is 1.8 x 10^-5. At one-half of the equivalence point, the concentration of CH3COOH is equal to the concentration of CH3COO-. Using the Henderson-Hasselbalch equation, we can calculate the pH:

pH = pKa + log ([CH3COO-] / [CH3COOH])

Substituting the values:

pH = -log(1.8 x 10^-5) + log (0.00125 mol / 0.00125 mol) = 4.74