Generate the titration curve of SB VS WA WACH COOH SB - Nant 5 = 0.02M Ka = 1.8×10 Vol = 50mL Concn = 0.02M Спаси conch = 0.02M Vadded, 0, 5, 10, 15, 20, 25, 40, 25, 40, 45, 50 & 55 PH = ?
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- Sketch two pH curves, one for the titration of a weak acid with a strong base and one for a strong acid with a strong base. How are they similar? How are they different? Account for the similarities and the differences.A 10.00-g sample of the ionic compound NaA, where A is the anion of a weak acid, was dissolved in enough water to make 100.0 mL of solution and was then titrated with 0.100 M HCl. After 500.0 mL HCl was added, the pH was 5.00. The experimenter found that 1.00 L of 0.100 M HCl was required to reach the stoichiometric point of the titration. a. What is the molar mass of NaA? b. Calculate the pH of the solution at the stoichiometric point of the titration.Sketch a pH curve for the titration of a weak acid (HA) with a strong base (NaOH). List the major species, and explain how you would go about calculating the pH of the solution at various points, including the halfway point and the equivalence point.
- Which of the indicators in Fig. 14-8 could be used for the titrations in Exercises 61 and 63?Prepared buffer solution GivenAmmonia Volume = 68 mL | Ammonia Concentration = 0.17 MAmmonium Chloride Volume = 42 mL | Ammonium Chloride Concentration = 0.13 MDissociation Constant of Ammonia: 1.8x10-5What is the total vol. of bufferConsider the titration of 50.0 mL of 0.0500 M HONH2 (a weak base; Kb = 1.10e-08) with 0.100 M HI. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mLpH = (b) 6.3 mLpH = (c) 12.5 mLpH = (d) 18.8 mLpH = (e) 25.0 mLpH = (f) 42.5 mLpH =
- OpenVellumHMAC=defce5f03204e95efa2fcf11b1ecff98#10001 I Review | Constants | Periodic Table [base] pH = pK, + log Tacid 3.5 x 10 2 -log(1.8 x 10-5)+ log 0.1025 4.74 – 0.467 4.28 Part C For 500.0 mL of a buffer solution that is 0.185 M in CH,CH,NH, and 0.135 M in CH3CH,NH CI, calculate the initial pH and the final pH after adding 2.0x10-2 mol of HCl (K,(CH CH,NH,) = 5.6 x 10 4) Express your answers using two decimal places separated by a comma. pHinitial. pHrinal 10.89, 11.12 Submit Previous Answers Request Answer X Incorrect; Try Again; 4 attempts remaining Provide Feedback Next > P Pearson at © 2022 Pearson Education Inc. All rights reserved. Terms of Use | Privacy Policy | Permissions Contact Us 11:57 PM 4/16/2022 lyp DDI C dele %3D backspace 00Consider the titration of 80.0 mL of 0.0200 M NH3 (a weak base; Kb = 1.80e-05) with 0.100 M HIO4. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mLpH = (b) 4.0 mLpH = (c) 8.0 mLpH = (d) 12.0 mLpH = (e) 16.0 mLpH = (f) 22.4 mLpH =4-Calculate pCl for the titration of 100.0 mL of 0.1000 M Cl¯ with 0.1000 M AGNO3 for the addition of 0.00 and 20.00 mL AGNO3. ptor TOCLR)3(AOYes) SHIFT WODE (SETUP)(1 MtOOMatho) AWA SETUPIOaOaLineo) A SETUPIELinelo) ETUPIOO4CONT) g F9 CLR) O(Setup)O(Yes)
- 5. In the titration of 15.00 mL of 0.200 M of a weak base, B:, (Kb = 7.25 x 10³) with 0.100 M HCI. calculate the pH after the addition of the following volumes of titrant (mL); (a) 0.00 ; (b) 10.00: (c ) 15.00 ; (d) 20.00 mL. %3DConsider the titration of 40.0 mL of 0.0600 M HONH2 (a weak base; Kb = 1.10e-08) with 0.100 M HI. Calculate the pH after the following volumes of titrant have been added: (d) 18.0 mLpH = (e) 24.0 mLpH = (f) 38.4 mLpH =50.0 mL of 0.0500 M HONH2 (a weak base; Kb = 1.10e-08) with 0.100 M HI. Calculate the pH after the following volumes of titrant have been added: (d) 18.8 mLpH = (e) 25.0 mLpH = (f) 42.5 mLpH =