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added to 25.00 mL of a 0.100 M HA solution in an acid-base neutralization reaction. M(OH)2 is a strong base where M2+ is a metal cation and OH– is the hydroxide ion. MM = 42.05 g/mol. HA is a strong acid where H+ is a proton and A– is an anion. MM = 65.23 g/mol.   A. Write and balance the chemical equation for the double replacement reaction (no state symbols needed). B. Determine the limiting reactant using moles of product and/or reactant. C. Calculate the moles of excess

added to 25.00 mL of a 0.100 M HA solution in an acid-base neutralization reaction. M(OH)2 is a strong base where M2+ is a metal cation and OH– is the hydroxide ion. MM = 42.05 g/mol. HA is a strong acid where H+ is a proton and A– is an anion. MM = 65.23 g/mol.   A. Write and balance the chemical equation for the double replacement reaction (no state symbols needed). B. Determine the limiting reactant using moles of product and/or reactant. C. Calculate the moles of excess

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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16.00 mL of a 0.125 M M(OH)2 solution is added to 25.00 mL of a 0.100 M HA solution in an acid-base neutralization reaction.

  • M(OH)2 is a strong base where M2+ is a metal cation and OH– is the hydroxide ion. MM = 42.05 g/mol.
  • HA is a strong acid where H+ is a proton and A– is an anion. MM = 65.23 g/mol.

 

A. Write and balance the chemical equation for the double replacement reaction (no state symbols needed).

B. Determine the limiting reactant using moles of product and/or reactant.

C. Calculate the moles of excess reactant left over. Assume the reaction goes to completion.

D. Calculate the pH of the final solution. You can assume volumes are additive for simplicity.

A. The balanced chemical reachion is given belows M(OH)2 t 2 HA MA2 + 2H20 1 mol 2mal 1mol B. 16 mL of O.125 M MCOH)2 ニ 012ら ×16 ニ 2 mmol of M(OH)2 25 m2 of 01 M HA = 25X 01 2'5 mmol of HA HA 50, eot is me limining reagent, The exess real tant deft over う 2.5 0.75 mmol ,ot MOH), 2 - ニ ニ 16 + 25 = To tal v ume of me soluh'om 0.15 XY 41 mL D. of föH] = 3,75 X 10 3 M lone. 30, the 41 Thus, 14+ log [on] = 14+ log (317SXI0-3) 11.57 (Ans).
Transcribed Image Text:A. The balanced chemical reachion is given belows M(OH)2 t 2 HA MA2 + 2H20 1 mol 2mal 1mol B. 16 mL of O.125 M MCOH)2 ニ 012ら ×16 ニ 2 mmol of M(OH)2 25 m2 of 01 M HA = 25X 01 2'5 mmol of HA HA 50, eot is me limining reagent, The exess real tant deft over う 2.5 0.75 mmol ,ot MOH), 2 - ニ ニ 16 + 25 = To tal v ume of me soluh'om 0.15 XY 41 mL D. of föH] = 3,75 X 10 3 M lone. 30, the 41 Thus, 14+ log [on] = 14+ log (317SXI0-3) 11.57 (Ans).
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