Construct a graph of pAg versus milliliters of Ag+ for the titration of 40.00 mL
ID: 1069036 • 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
trimethyl amine = Kb = 2.6 x 10^-6
pKb = 5.58
millimoles of B = 100 x 0.1 = 10
(a) initial pH
pOH = 1/2 [pKb- logC]
pOH = 1/2 [5.58 - log 0.1]
= 1/2 (5.58 - log 0.1)
= 3.29
pH + pOH = 14
pH = 14 - pOH
= 10.71
pH = 10.71
b) 20 mL titrant added :
millimoles of HCl = 20 x 0.2 = 4
B + H+ --------------------> BH+
10 4 0
6 0 4
pOH = pKb + log [salt / base]
= 5.58 + log [4 / 6]
= 5.40
pH = 8.60
c) 49 mL
millimoles of HCl = 49 x 0.2 = 9.8
B + H+ --------------------> BH+
10 9.8 0
0.2 0 9.8
pOH = pKb + log [salt / base]
= 5.58 + log [9.8 / 0.2]
= 7.27
pH = 6.73
d) 50 mL
millimoles of HCl = 50 x 0.2 = 10
B + H+ --------------------> BH+
10 10 0
0 0 10
salt concentration = 10/ (100 +50 ) = 0.067 M
pH = 7 - 1/2 [pKb + logC]
pH = 7 - 1/2 [5.58 + log 0.067]
pH = 4.80
e) 50.1
millimoles of acid = 10.02
strong acid millimoles remains = 10.02 - 10 = 0.02
strong acid concentrtion = 0.02 / (100 + 50.1) = 1.33 x 10^-4 M
pH = -log [H+]
pH = -log (1.33 x 10^-4)
pH = 3.88
note : similarly solve remaining points like part e)