Chapter 7: Let the Titrations Begin 10. 0.100 M Pb2+. Ksp = 7.4 x 10-14 for...
Ksp for SrSO4 = 3.2 x 10^-7 A solution contains 0.0210 M Pb2 + (aq) and 0.0210 M Sr2 + (aq). If you add so - (aq), what will be the concentration of Pb2 + (aq) when Srso,() begins to precipitate? [Pb2+] = M
а он (15 points) A 0.100 M oxalic acid, HO,CCOH, solution is titrated with 0.100 M KOH. Calculate the pHs when 25.00 mL of oxalic acid solution is titrated with 10, 15, 20, 25, 35, 40, 45, 50 and 55 mL of Nont Kal = 5.4 x 10-2 and K 2 = 5.42 x 10- for oxalic acid. Show all your work and make a graph of pH versus Vo HO, CCO 2H + KOH KC204 + H2Oenwot aunt men...
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....
X) The answers: a) 0.315 V c) 0.273 V d) 1.80 M 10. For the voltaic cell: Cd(s) Cd2 (0.100 M, 25 mL) || Pb2 (2.000 M, 25 mL) | Pb(s) a. What is Eel initially? (E 0.277 v) b. If the cell is allowed to operate spontaneously, will Eel increase, decrease or remain constant with time? Explain. c. What will be Ee when [Pb2"] has fallen to 0.900 M? [Hint: Both concentrations change. Why?] d. What will be [Cd2"]...
What is the pH of a solution of 40.0 mL of 0.100 M acetic acid (Ka = 1.8 x 10-5) after 50.0 mL of 0.100 M NaOH has been added? Calculate the concentration of dissolved Ba2+ ions when BaSO4 is added to water at 25°C. Кsp? = 1.10 x 10-10 A particular saturated solution of silver chromate (Ag2CrO4), has [Ag+] = 5.0 x 10 Mand (CrO4) = 4.4 x 10M. What is value Ksp for silver chromate? As a result...
7. Calculate the pH of the solution obtained by titrating 50.0 mL of 0.100 M HNO2(aq) with 0.150 M NaOH(aq) to the equivalence point. Take Ka = 5.6 x 10 - M for HNO2(aq).
Please help with d-g TITRATIONS CHEM 2. 20.00 mL of 0.270 M HA (K. = 7.2 x 10-4) was titrated with 0.300 M sodium hydroxide a) Write the titration equation and calculate volume of NaOH needed to reach the equivalence point. mL b) At what volume of the titrant would pH=pKa? Why? c) pH at the equivalence point when the acid and base are neutralized will be -. choose one of these answer choices given below: i) equal to 7...
Weak-Acid Strong-Base Titrations. These next questions relate to a 25 mL aliquot of 0.35 M acetic acid (Ka = 1.77 x 10) that is titrated with 0.20 M potassium hydroxide (KOH). (f) What is the pH of the acetic acid solution before the titration begins? (g) What is the pH after 14 mL of 0.20 M KOH has been added to the solution? Use the Henderson-Hasselbalch equation. (h) What is the pH at the equivalence point?
Nicole measured the pH of a 0.100 M solution of propanoic acid, CH3CH2COOH, a weak organic acid at equilibrium and found it to be 2.931 at 25C. Calculate the Ka of propanoic acid. Her lab instructor mentions that the half equivalence method is better for determining pKa. What is the half-equivalence point and why is this method better at determining pKa? What is the pH after 20.0 mL of 0.0750 M NaOH is added to 30.0 mL of the 0.100...
(10 marks) Calculate the pH after titrating 50.0 mL of a 0.100 M weak base solution (Kb = 7 x10-9) with: 0.200 M HBr to the equivalence point b. 50.0 mL of 0.200 M HBr