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Answer questions 1 c, d and e only!

Answer questions 1 c, d and e only!

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter14: Chemical Equilibrium
Section: Chapter Questions
Problem 14.74QE: Lead poisoning has been a hazard for centuries. Some scholars believe that the decline of the Roman...
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Answer questions 1 c, d and e only!
3:18 PM Wed Nov 4 * 43% + : O Using the mmoles of salt produced and the total volume in ml of solution (from the mixing of the acid and base solutions), calculate the molarity of the salt solution produced by this reaction. (e) Consider this example problem: If 100. ml of 0.100 M HCI solution is mixed with 100. ml of 0.100 M NaOH, what is the molarity of the resulting salt solution? (assuming the volumes are additive and ignore the change in H20, which is negligible). HCla) NaOHagi Nacla) 1 mol 1 mol 1 mol 1 mol Rxn ratio: 100 ml x (0.100 mmol NCI 1 ml 100 ml x (0.100 mmol NaoN 1 ml. Mols @ Start: O mol -- = 10 mmol HBr = 10 mmol NaOH Change -10 mmol - 10 mmol NaOH + 10 mmol O mmol O mmol After rxn 10 mmol The resulting solution contains 10 mmol NaCl in a total volume of 200 mL. The molarity of NaCl solution is 10 ттol = 0.05 M NaCl 200 mL Fill in a table like this for the calculations in problem (1) (a)-(d). 1 4
Transcribed Image Text:3:18 PM Wed Nov 4 * 43% + : O Using the mmoles of salt produced and the total volume in ml of solution (from the mixing of the acid and base solutions), calculate the molarity of the salt solution produced by this reaction. (e) Consider this example problem: If 100. ml of 0.100 M HCI solution is mixed with 100. ml of 0.100 M NaOH, what is the molarity of the resulting salt solution? (assuming the volumes are additive and ignore the change in H20, which is negligible). HCla) NaOHagi Nacla) 1 mol 1 mol 1 mol 1 mol Rxn ratio: 100 ml x (0.100 mmol NCI 1 ml 100 ml x (0.100 mmol NaoN 1 ml. Mols @ Start: O mol -- = 10 mmol HBr = 10 mmol NaOH Change -10 mmol - 10 mmol NaOH + 10 mmol O mmol O mmol After rxn 10 mmol The resulting solution contains 10 mmol NaCl in a total volume of 200 mL. The molarity of NaCl solution is 10 ттol = 0.05 M NaCl 200 mL Fill in a table like this for the calculations in problem (1) (a)-(d). 1 4
3:34 PM Wed Nov 4 * 36% I Given the balancedequation: HBr(aq) + NAOH(aq) → NaBr(aq) + H2OU» and remembering that Molarity = moles/liter OR mmoles/mL (1) (a) Calculate the number of mmoles of HBr in 100.0 ml of 0.250 MHBR mmoles of HBr = molarity x volume O-250 x 100 205 mmoles HBr (6) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 MNAOH mmoles of NaOH = molo Lolarity olume 0.250 x l00 %3D 2.5 mmoles NaoH © When these two solutions are mixed the acid and pase snouid neutralize one another exactly. This means that all of the acid and base are completely used up; either one could be considered a "limiting reactant". Starting with the mmoles of either the acid base, calculate number of mmoles of salt produced by the reaction. (d) Using the mmoles of salt produced and the total volume in mL of solution (from the mixing of the acid and base solutions), calculate the molarity of the salt solution produced by this reaction. 1 4 (e) Consider this example problem: If 100. ml of 0.100 M HCI solution is mixed with 100. ml of 0.100 M NaOH, what is the molarity of the resulting salt solution? (assuming the volumes are additive and ignore the change in H20, which is negligible). HClaa) NaOHjaal Naclja) I mol 1 mol 1 mol 1 mol Ryn ratio:
Transcribed Image Text:3:34 PM Wed Nov 4 * 36% I Given the balancedequation: HBr(aq) + NAOH(aq) → NaBr(aq) + H2OU» and remembering that Molarity = moles/liter OR mmoles/mL (1) (a) Calculate the number of mmoles of HBr in 100.0 ml of 0.250 MHBR mmoles of HBr = molarity x volume O-250 x 100 205 mmoles HBr (6) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 MNAOH mmoles of NaOH = molo Lolarity olume 0.250 x l00 %3D 2.5 mmoles NaoH © When these two solutions are mixed the acid and pase snouid neutralize one another exactly. This means that all of the acid and base are completely used up; either one could be considered a "limiting reactant". Starting with the mmoles of either the acid base, calculate number of mmoles of salt produced by the reaction. (d) Using the mmoles of salt produced and the total volume in mL of solution (from the mixing of the acid and base solutions), calculate the molarity of the salt solution produced by this reaction. 1 4 (e) Consider this example problem: If 100. ml of 0.100 M HCI solution is mixed with 100. ml of 0.100 M NaOH, what is the molarity of the resulting salt solution? (assuming the volumes are additive and ignore the change in H20, which is negligible). HClaa) NaOHjaal Naclja) I mol 1 mol 1 mol 1 mol Ryn ratio:
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