18.0-mL sample of 0.126 M diprotic acid (H2A) solution is titrated with 0.1040 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5 and Ka2=3.4×10−10.
Part A
At what added volume of base does the first equivalence point occur?
Part B
At what added volume of base does the second equivalence point occur?
18.0-mL sample of 0.126 M diprotic acid (H2A) solution is titrated with 0.1040 M KOH. The...
22.0-mL sample of 0.122 M diprotic acid (H2A) solution is titrated with 0.1016 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5 and Ka2=3.4×10−10. A. At what added volume of base does the first equivalence point occur? ____ mL B. At what added volume of base does the second equivalence point occur? _____ mL 12
20.0-mL sample of 0.123 M diprotic acid (H2A) solution is titrated with 0.1013 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5 and Ka2=3.4×10−10. Part A: At what added volume of base does the first equivalence point occur? Part B: At what added volume of base does the second equivalence point occur?
Review I Constants I Periodic Table 20.0-mL sample of 0.123 M diprotic acid (H2 A) solution is titrated with 0.1021 M KOH. The acid ionization constants for the acid are Part A Ka1 5.2 x 10 and Ka, =3.4 x 10-10. At what added volume of base does the first equivalence point occur? η ΑΣφ V= 21.7 mL Previous Answers Request Answer Submit X Incorrect; Try Again; 14 attempts remaining Part B At what added volume of base does the...
6. A student titrated 50.0 mL of the 0.10 M unknown diprotic H2A with 0. 10 M NAOH. After 25.0 mL of NaOH was added, the pH of the resulting solution was 6.70. After 50.0 mL of NaOH was added, the pH of the solution was 8.00. What are the values of Ka1 and Ka2? 6. A student titrated 50.0 mL of the 0.10 M unknown diprotic H2A with 0. 10 M NAOH. After 25.0 mL of NaOH was added,...
Acid/Base titrations 10. 25.0 mL of 0.100 M H2A (a weak diprotic acid) is titrated with 0.200 M NaOH. What is the pH of the solution when 0.00 mL, 10.0 mL, 12.5 mL, 20.0 mL, 25.0 mL, and 40.0 mL E.S RaWeMA 5.83 x 10 8) have been added? (Ka1= 2.46 x 10, Ka2 ANSWER: 0 mL = 2.315, 10.0 mL= 4.211 (or 4.213), 12.5 mL = 5.422, 20.0 mL 7.410, 25.0 mL = 9.966, 40.0 mL = 12.664 11....
Given a diprotic acid, H2A, with two ionization constants of Ka1 = 3.5× 10–4 and Ka2 = 5.7× 10–12, calculate the pH for a 0.113 M solution of NaHA. PH=
For the diprotic weak acid H2A, Ka1 = 2.7 × 10-5 and Ka2 = 5.2 × 10-7. What is the pH of a 0.0550 M solution of H2A? What are the equilibrium concentrations of H2A and A2– in this solution?
ration 46. A 30.0-ml sample of 0.165 M propanoic acid is titrated with 0.300 M KOH. Calculate the pH at each volume of added base: OmL, 5 mL, 10 mL, equivalence point, one-half equivalence point. 20 mL, 25 mL. Use your calculations to make a sketch of the titration curve. ed by
A diprotic acid, H2A, has Ka1 = 3.4 x 10-4 and Ka2 = 6.7 x 10-9. What is the pH of a 0.36 M. solution of H2A?
A 100.0-mL aliquot of 0.100 M diprotic acid H2A (PK1 = 4.00, PKa2 = 8.00) was titrated with 1.00 M NaOH. Find the pH at the following volumes of base added. a) 1 ml, b) 11 mL, c) 20 mL and d) 22 mL.