A 1000 mL aliquot of .1 M diprotic acid H2A (pK1=4; pK2=8) was titrated with 1 M NaOH. find pH values at Vb: 0, 3, 8, 10, 13, 19, 20, 22.
A 1000 mL aliquot of .1 M diprotic acid H2A (pK1=4; pK2=8) was titrated with 1...
A 100.0 mL aliquot of 0.100M diprotic acid H2A(pK1 = 4.00; pK2 = 8.00) was titrated with 1.00 M NaOH. Find the pH values at the following volumes of base added, Vb; Vb= 0 mL, PH= Vb= 2 mL, PH= Vb= 8 mL, PH= Vb= 10 mL, PH= Vb= 12 mL, PH= Vb= 18 mL, PH= Vb= 20 mL, PH= Vb= 22 mL, PH=
A 125.0-mL aliquot of 0.121 M diprotic acid H2A (pK1 = 4.01, pK2 = 8.02) was titrated with 1.21 M NaOH. Find the pH at the following volumes of base added: Vb = 12.50, 13.50, 18.75, 24.00, 25.00, and 29.00 mL. (Assume Kw = 1.01 ✕ 10−14.)
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.
A 100.0 mL aliquot of 0.100 M monoprotic acid H2A (pK1 = 4.00) was titrated with 1.00 M NaOH. Find the pH at the following volumes of base added: Vb = 0, 5, 9, 10 and 11 mL. Sketch the titration curve.
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,...
A 100.0m aliquot of 0.100 M diprotic acid H₂A (pk, = 4.00; ph2 = 8.00) was titrated with 1.00M NaOH. Find the pH at the following volumes of base added, Vb: 0mL, 2mL, 8mL, 10mL, 12 mL, 18mL, 20mL, 22mL,
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....
A student titrated a 25.00-mL sample of a solution containing an unknown weak, diprotic acid (H2A) with NaOH. If the titration required 17.73 mL of 0.1036 M NaOH to completely neutralize the acid, calculate the concentration (in M) of the weak acid in the sample. (a) 9.184 x 10‒4 M (b) 3.674 x 10‒2 M (c) 7.304 x 10‒2 M (d) 7.347 x 10‒2 M (e) 1.469 x 10‒1 M
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
1. Titration of a diprotic acid, H2A. Consider the titration of 50 mL of 0.02 M H2A with 0.1 M NaOH. pKa1 = 4.00 and pKa2 = 8.00. For each point in the titration, calculate [H+]. a) Before any titrant is added b) after 5 mL of titrant added c) after 6 mL titrant is added d) after 10 mL titrant is added e) after 15 mL titrant is added f) after 17 mL titrant is added g) after 20...