Consider the titration of a weak acid that has a pKa = 4.00. Suppose a chemist was going to perform a titration on 50.0 mL of 0.050 0 M of the weak acid using 0.500 M NaOH. (A) What would be the pH of the solution after 0.00 mL of 0.500 M NaOH has been added?(B) What would be the pH halfway to the equivalence point of the titration?(C) What would be the pH at the equivalence point in the titration? (D) What would be the pH of the solution after 6.00 mL of 0.500 M NaOH has been added?
Consider the titration of a weak acid that has a pKa = 4.00. Suppose a chemist was going to perform a titration on 50.0 mL of 0.050 0 M of the weak acid using 0.500 M NaOH. (A) What would be the pH of the solution after 0.00 mL of 0.500 M NaOH has been added?(B) What would be the pH halfway to the equivalence point of the titration?(C) What would be the pH at the equivalence point in the titration? (D) What would be the pH of the solution after 6.00 mL of 0.500 M NaOH has been added?
Use the following reasoning when solving this problem.
Because at the equivalence point the moles of strong base (0.500 M NaOH) added as a titrant equal the moles of weak acid, HA (50.0 ml of 0.050 M) in the solution, we can start by calculating the volume of 0.500 M NaOH needed to reach the equivalence point. To do this we simply equate the moles of HA to moles of NaOH by using the dilution formula. i.e. MaVa = MbVb where a represents HA and b represents NaOH. To calculate volume of NaOH needed to reach the equivalence point, Vb, we plug in the molarities for HA (Ma) and NaOH (Mb) and volume of HA (Va) in the formula and solve for Vb.Use the obtained value of Vb to determine which point of the titration each of the 4 given volumes represents. You should find the value of Vb to be .... ml).
(a) 0.00 mL added NaOH, is the region before any titrant is added. pH is determined by the weak acid equilibrium constant equation (HA + H2O ↔ H3O+ + A-). Use the Ka expression to determine the pH.
(b) If Vb if greater than 2.5 mL added NaOH - This is the region beforethe equivalence point. HA is partially converted to A-, so we have a mixture of HA and A- (A buffer). We use the Henderson-Hasselbalch Eq. to find pH. Subtract moles of added NaOH from initial moles of HA to calculate remaining moles of HA. The amount of conjugate base A- formed is equal to the moles of added NaOH.
(c) If Vb is equal to 5 mL added NaOH, The titration is at the equivalence point: “all” HA has been converted to A-, the conjugate weak base. pH is determined from (A- + H2O ↔ HA + OH-), so we have to use the Kb expression to determine the pH. The concentration of A- is based of the original concentration of HA times the dilution factor (initial volume/total volume).
(d) If 6 mL added NaOH is greater than Vb, This is the region after the equivalence point: All HA has been converted to A- and we have excess OH- from the strong base. pH determined by excess OH- since OH- is a much stronger base than A-. Subtract moles of HA from moles of added NaOH to find moles of excess OH- and divide the obtained value of moles by the total volume of solution (50.0 mL + 6.00 mL = 56.00 mL = 0.056 L) to get the molarity of excess OH-, then the pOH and pH.
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