Please explain the result for letter D because I don't understand 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.
Please explain the result for letter D because I don't understand 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.
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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Please explain the result for letter D because I don't understand
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).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06f38b42-94e8-415c-9731-0f8a1da31895%2F61e7e9d3-f1d6-4d7b-a344-6d0fea9753d4%2Fiak4qvk_processed.jpeg&w=3840&q=75)
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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