Include an annotated calculation showing how you determined the Kai and Ka2 (i.e. How did you find Ka1 and Ka2?). Provide a percent error comparing your experimental molar mass to the accepted value. 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} ]
Include an annotated calculation showing how you determined the Kai and Ka2 (i.e. How did you find Ka1 and Ka2?). Provide a percent error comparing your experimental molar mass to the accepted value. 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} ]
Chapter9: Acids, Bases, And Salts
Section: Chapter Questions
Problem 9.109E
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Transcribed Image Text:Include an annotated calculation showing how you determined the Kai and Ka2 (i.e. How
did you find Ka1 and Ka2?).
Provide a percent error comparing your experimental molar mass to the
accepted value.
![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} ]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4c9398a2-c49f-444d-8855-e495f02ff958%2Fde28bdb9-372a-4978-a93c-c96ebfef86fc%2Flr4aq4c_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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} ]
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