1. Determining the Molar Mass of the Acid • Given mass of acid: 0.120 g • Volume of NaOH solution at first • equivalence point: 12.5 mL Molarity of NaOH: 0.1 M Calculation of moles of NaOH: [ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.1, M \times 0.0125, L = 1.25 \times 10^{-3}, \text{mol} ] Moles of Acid: Since the acid is diprotic, the moles of acid will be half the moles of NaOH used: [ \text{Moles of Acid} = \frac{1.25 \times 10^{-3}, \text{mol}}{2} = 6.25 \times 10^{-4}, \text{mol} ] Molar Mass of the Acid: [ \text{Molar Mass} \frac{\text{Mass of Acid}} = {\text{Moles of Acid}} = \frac{0.120, \text{g}}{6.25 \times 10^{-4}, \text{mol}} = 192, \text{g/mol} ] 2. Determining (K_a1) and (K_a2) • ⚫ pH at the first half-equivalence point (pH 1st half): 2.91 ⚫ pH at the second half-equivalence point (pH 2nd half): 7.5 Since pKa = pH at half equivalence point: (K_{a1}): [pka_1 = 2.91 ] [ K_{a1} = 10^{-2.91} \approx 1.23 \times 10^{-3}] (K_{a2}): [ pKa_2 = 7.5 ] [ K_{a2} = = 10^{-7.5} \approx 3.16 \times 10^{-8} ]

1. Determining the Molar Mass of the Acid • Given mass of acid: 0.120 g • Volume of NaOH solution at first • equivalence point: 12.5 mL Molarity of NaOH: 0.1 M Calculation of moles of NaOH: [ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume} = 0.1, M \times 0.0125, L = 1.25 \times 10^{-3}, \text{mol} ] Moles of Acid: Since the acid is diprotic, the moles of acid will be half the moles of NaOH used: [ \text{Moles of Acid} = \frac{1.25 \times 10^{-3}, \text{mol}}{2} = 6.25 \times 10^{-4}, \text{mol} ] Molar Mass of the Acid: [ \text{Molar Mass} \frac{\text{Mass of Acid}} = {\text{Moles of Acid}} = \frac{0.120, \text{g}}{6.25 \times 10^{-4}, \text{mol}} = 192, \text{g/mol} ] 2. Determining (K_a1) and (K_a2) • ⚫ pH at the first half-equivalence point (pH 1st half): 2.91 ⚫ pH at the second half-equivalence point (pH 2nd half): 7.5 Since pKa = pH at half equivalence point: (K_{a1}): [pka_1 = 2.91 ] [ K_{a1} = 10^{-2.91} \approx 1.23 \times 10^{-3}] (K_{a2}): [ pKa_2 = 7.5 ] [ K_{a2} = = 10^{-7.5} \approx 3.16 \times 10^{-8} ]

Introductory Chemistry For Today

8th Edition

ISBN:9781285644561

Author:Seager

Publisher:Seager

Chapter9: Acids, Bases, And Salts

Section: Chapter Questions

Problem 9.109E

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