The graph below shows the pH changes during the titration of a weak acid with so

Question

The graph below shows the pH changes during the titration of a weak acid with sodium hydroxide (NaOH). Based on the titration curve, determine the pK_a of the acid. Find the half-equivalence volume on the titration curve. The half-equivalence point is the point at which half the volume of base required to reach equivalence has been added. For the titration shown, mL of NaOH is required to neutralize the acid and reach equivalence. Therefore, the half-equivalence point occurs when half this volume of base, mL, has been added. Find the pH at the half-equivalence point. (Enter your answer to one decimal place to account for the precision of the values on the x and y-axes.) Find the pH of the titration solution at the half-equivalence point. Trace a vertical line from the 15 mL point on the x-axis up to the curve. From there, trace a horizontal line to the y-axis to find the pH, which is approximately. At the half-equivalence point, the pH is equal to the pK_a of the acid. This can be shown by examining the Henderson-Hasselbalch equation and the dissociation of the acid. At the half-equivalence point, the concentrations of dissociated acid (A) and undissociated acid (HA) are equal. Therefore, the ratio of their concentrations is 1 and the logarithm of 1 is 0, eliminating the right-hand term in the equation. This leaves pH = pKa. HA + H_2O H_3O^+ + A^- pH = pK_a + log [A^-]/[HA] Therefore, the pH at the half-equivalence point is equal to the pK_a of the acid, which is -

Explanation / Answer

For the titration shown, 30mL of NaOH is required to neutralize the acid and reach the equivalence. Therefore, the half equivalence point occurs when half of this volume of base, 15mL, has been added.

Step2

The pH from the vertical line was found to be approximately 3.8

At the haf equivalence point, the pH equals to the pKa of the acid. The pH is determined using Hendersen-Hasselbalch equation,

pH = pKa + log ([A-]/[HA])

At the half equivalence pont, [A-] = [HA], results in log1 which equals to zero

pH = pKa

There fore, the pH at the half equivalence point equals to the pKa of the acid, which is 3.8.

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