A 50.0-mL solution of the acid H₃A (K_a1 = 1.0 × 10^-6 , K_a2 = 1.0 × 10^-9, and K_a3 = 1.0 × 10^-12.) is titrated with 0.050 M KOH. The second equivalence point is reached at 25.0 mL of base. Calculate the original concentration of H₃A and the volume of 0.050 M KOH needed to reach a pH of 6.70.

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The Answer of A 50.0-mL solution of the acid H₃A (K_a1 = 1.0 × 10^-6 , K_a2 = 1.0 × 10^-9, and K_a3 = 1.0 × 10^-12.) is titrated with 0.050 M KOH. The second equivalence point is reached at 25.0 mL of base. Calculate the original concentration of H₃A and the volume of 0.050 M KOH needed to reach a pH of 6.70.

A 500-ML Solution of the Acid H3A (Ka1 = 1

A 500-ML Solution of the Acid H₃A (K_a1 = 1