Construct a graph of pAg versus milliliters of Ag+ for the titration of 40.00 mL
ID: 1069034 • Letter: C
Question
Construct a graph of pAg versus milliliters of Ag+ for the titration of 40.00 mL of solution containing 0.050 M Br^- and 0.050 M CI^- The titrant is 0.08454 M AgNO_3. Calculate pAg^+ at the following volumes: 2.00. 10.00. 22.00. 23.00, 24 00, 30.00, 40.00 mL, second equivalence point 50 mL. Ksp (AgBr) = 5.0 times 10^-13 Ksp (AgCl) = 1.8 times 10^-10 Calculate the pH at each point listed for the titration of 100.00 mL of 0.1 M cocaine [K_b = 2.6 times 10^-6] with 0.2 M HNO_3. The points to calculate are Va = 0.0, 20.0, 49.0, 50.0. 50.1, 51.0. an 60.0 mL. B + H = BH^- A 0.1 M solution of the weak acid HA was titrated with 0.1 M NaOH. The pH measured when Vbase = 1/2 V_eq.point was 4.62. Using activity coefficients, calculate pKa.f_A = 0.854, f_HA = 1.00 A sample of impure salicylic acid, C_6H_4 (OH) COOH [Mwt:138 mg/mmol] (one titratabl proton), is analyzed by titration. What size sample should be taken so that the percent purity equal to five times the milliliters of 0.05 M NaOH used to titrate it? When 100 0 mL of a weak acid was titrated with 0.09381 M NaOH, 27.63 mL were required to reach the equivalence point. The pH at the equivalence point was 10.99. What was the when on ly 19.47 mL of NaOH had been added?Explanation / Answer
1. Titration
(a) 2 ml AgNO3 added
moles of Br- = 0.05 x 40 = 2 mmol
moles of AgNO3 = 0.08454 x 2 = 0.16908 mmol
excess [Br-] = 1.83092 mmol/42 ml = 0.0436 M
[Ag+] = 5 x 10^-13/0.0436 = 1.147 x 10^-11 M
E = 0.7999 + 0.0592 log(1.147 x 10^-11) = 0.152 V
(b) 10 ml AgNO3 added
moles of Br- = 0.05 x 40 = 2 mmol
moles of AgNO3 = 0.08454 x 10 = 0.8454 mmol
excess [Br-] = 1.1546 mmol/50 ml = 0.023092 M
[Ag+] = 5 x 10^-13/0.023092 = 2.165 x 10^-11 M
E = 0.7999 + 0.0592 log(2.165 x 10^-11) = 0.168 V
(c) 22 ml AgNO3 added
moles of Br- = 0.05 x 40 = 2 mmol
moles of AgNO3 = 0.08454 x 22 = 1.85988 mmol
excess [Br-] = 0.14012 mmol/62 ml = 0.00226 M
[Ag+] = 5 x 10^-13/0.0026 = 2.21 x 10^-10 M
E = 0.7999 + 0.0592 log(2.21 x 10^-10) = 0.228 V
(d) 23 ml AgNO3 added
moles of Br- = 0.05 x 40 = 2 mmol
moles of AgNO3 = 0.08454 x 23 = 1.94442 mmol
excess [Br-] = 0.05558 mmol/63 ml = 0.00088 M
[Ag+] = 5 x 10^-13/0.0088 = 5.67 x 10^-10 M
E = 0.7999 + 0.0592 log(5.67 x 10^-10) = 0.619 V
(e) 24 ml AgNO3 added
moles of Cl- = 0.05 x 40 = 2 mmol
moles of AgNO3 = 0.08454 x 24 = 2.02896 mmol
excess [Ag+] = 0.02896 mmol/64 ml = 0.0004525 M
E = 0.7999 + 0.0592 log(0.0004525) = 0.602 V
(f) 30 ml AgNO3 added
moles of Cl- = 0.05 x 40 = 2 mmol
moles of AgNO3 = 0.08454 x 30 = 2.5362 mmol
excess [Ag+] = 0.5362 mmol/70 ml = 0.00766 M
E = 0.7999 + 0.0592 log(0.00766) = 0.675
(g) 40 ml AgNO3 added
moles of Cl- = 0.05 x 40 = 2 mmol
moles of AgNO3 = 0.08454 x 40 = 3.3816 mmol
excess [Cl-] = 1.3816 mmol/80 ml = 0.01727 M
E = 0.7999 + 0.0592 log(0.01727) = 0.695 V
Similarly the other values can be calculated