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The following information is for questions 1 and 2. Consider the titration of 25.0 mL of 0.100 MHCI with 0.100 M NaOH. Calculate the pH after adding the following amounts of the NaOH solution: 1. 22.0 mL a) 1.27 b) 1.00 2.19 d) 3.52 e) 1.92

The following information is for questions 1 and 2. Consider the titration of 25.0 mL of 0.100 MHCI with 0.100 M NaOH. Calculate the pH after adding the following amounts of the NaOH solution: 1. 22.0 mL a) 1.27 b) 1.00 2.19 d) 3.52 e) 1.92

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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can someone explain what I did wrong. the correct answer is circled for number one, but I got 2.19 as the poh, not the pH

- NaOH + HC) ⇒ NaCl + H₂0 Dinitial moles of NaOH: 7500 × 0.1 = 0.0025 mol 1000 (a) moles of NaOH added: 22 x 0.1 = 0.0022 mal 1000 excess moles of NaOH: G 0.0025-0.0022 = 0.0003 mol Total volume = 25 mL + 22 = 47mL 0.047 L -3 COH-] = exass mole - 0.0003 6013 X10 total val 0.047 рон pOH = -log (OH-] = 10g [613X10²³] = 2.19 -p+) = 14-pOH = 14-2.19 - 11.81 XID
Transcribed Image Text:- NaOH + HC) ⇒ NaCl + H₂0 Dinitial moles of NaOH: 7500 × 0.1 = 0.0025 mol 1000 (a) moles of NaOH added: 22 x 0.1 = 0.0022 mal 1000 excess moles of NaOH: G 0.0025-0.0022 = 0.0003 mol Total volume = 25 mL + 22 = 47mL 0.047 L -3 COH-] = exass mole - 0.0003 6013 X10 total val 0.047 рон pOH = -log (OH-] = 10g [613X10²³] = 2.19 -p+) = 14-pOH = 14-2.19 - 11.81 XID
The following information is for questions 1 and 2. Consider the titration of 25.0 mL of 0.100 MHCI with 0.100 M NaOH. Calculate the pH after adding the following amounts of the NaOH solution: 1. 22.0 mL a) 1.27 b) 1.00 2.19 d) 3.52 e) 1.92 2. 25.0 mL a) 2.30 b) 1.00 c) 2.60 d) 1.30 7.00 lor folubility of BaF2 in pure water. (Ksp for BaF2=2.45 x 10) 577²
Transcribed Image Text:The following information is for questions 1 and 2. Consider the titration of 25.0 mL of 0.100 MHCI with 0.100 M NaOH. Calculate the pH after adding the following amounts of the NaOH solution: 1. 22.0 mL a) 1.27 b) 1.00 2.19 d) 3.52 e) 1.92 2. 25.0 mL a) 2.30 b) 1.00 c) 2.60 d) 1.30 7.00 lor folubility of BaF2 in pure water. (Ksp for BaF2=2.45 x 10) 577²
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