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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Question
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa389d800-2220-480d-b079-1654ac8487de%2Fa700dd81-38b7-44d5-bdc4-a3dd61cac2d9%2F9wpi9zb_processed.jpeg&w=3840&q=75)
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:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa389d800-2220-480d-b079-1654ac8487de%2Fa700dd81-38b7-44d5-bdc4-a3dd61cac2d9%2F1sghfh6_processed.jpeg&w=3840&q=75)
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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