A titration of 200.0 mL of 1.00 M H₂A was done with 1.02 M NaOH.For the diprotic acid H₂A,K_a1 = 2.5 $\times$10^-⁵,K_a2 = 3.1 $\times$ 10^-⁹.Calculate the pH after 600.0 mL of 1.02 M NaOH have been added.

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After adding 600.0 mL of 1.02 M NaOH to 200.0 mL of 1.00 M H₂A, the reaction goes beyond the equivalence point for both dissociations of the diprotic acid. The excess NaOH determines the pH. Calculate moles of NaOH added and moles of H₂A initially present. Excess moles of NaOH = total moles NaOH - 2 * moles H₂A. The concentration of OH- in the solution is then calculated from the excess moles of NaOH divided by the total volume of the solution. Finally, use pOH and the relationship pH + pOH = 14 to find the pH.

A Titration of 200 ×\times× 10-5,Ka2 = 3 ×\times× 10-9

A Titration of 200 $\times$ 10^-⁵,K_a2 = 3 $\times$ 10^-⁹