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k0ka [10]
3 years ago
11

An animal cell placed in a hypertonic solution will shrink in a process called crenation. An animal cell placed in a hypotonic s

olution will swell and potentially burst in a process called hemolysis. To prevent crenation or hemolysis, an animal cell must be placed in an isotonic solution such as 0.9% (m/v) NaCl or 5.0% (m/v) glucose. This does not mean that a cell has a 5.0% (m/v) glucose concentration; it just means that 5.0% (m/v) glucose will exert the same osmotic pressure as the solution inside the cell, which contains several different solutes. A red blood cell is placed into each of the following solutions. Indicate whether crenation, hemolysis, or neither will occur.
Solution A: 3.21% (m/v) NaCl
Solution B: 1.65% (m/v) glucose
Solution C: distilled H2O
Solution D: 6.97% (m/v) glucose
Solution E: 5.0% (m/v) glucose and 0.9%(m/v) NaCl
Drag each solution to the appropriate bin.
Chemistry
1 answer:
tresset_1 [31]3 years ago
3 0

Answer:

Solution A: 3.21% (m/v) NaCl  ⇒ CRENATION

Solution B: 1.65% (m/v) glucose  ⇒ HEMOLYSIS

Solution C: distilled H₂O  ⇒ HEMOLYSIS

Solution D: 6.97% (m/v) glucose  ⇒ CRENATION

Solution E: 5.0% (m/v) glucose and 0.9%(m/v) NaCl ⇒ CRENATION

Explanation:

Isotonic solution = 0.9% (m/v) NaCl or 5.0% (m/v) glucose

<u>Crenation</u> will occur if the solution has a concentration higher than 0.9% (m/v) NaCl or 5.0% (m/v) glucose (hypertonic solution). This will occur in the following solutions:

Solution A: 3.21% (m/v) NaCl  > 0.9% (m/v) NaCl

Solution D: 6.97% (m/v) glucose  > 5.0% (m/v) glucose

Solution E: 5.0% (m/v) glucose <u>and</u> 0.9%(m/v) NaCl (the addition of the 2 components exceeds the osmotic pressure permitted).

<u>Hemolysis</u> will occur if the solution has a concentration lower than 0.9% (m/v) NaCl or 5.0% (m/v) glucose (hypotonic solution). This is the case of:

Solution B: 1.65% (m/v) glucose < 5.0% (m/v) glucose

Solution C: distilled H₂O = 0% glucose/0% NaCl < 0.9% (m/v) NaCl or 5.0% (m/v) glucose

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Consider the titration of 82.0 mL of 0.140 M Ba(OH)2 by 0.560 M HCl. Calculate the pH of the resulting solution after the follow
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Answer:

a) pH = 13.447

b) pH = 13.176

Explanation:

Ba (OH)₂ is a strong base and ionizes completely in solution to give barium amend hydroxide ions.

The equation of the dissociation of Ba(OH)₂ is given below:

Ba(OH)₂ ----> Ba2+ + 2OH-

1 mole of Ba(OH)₂ produces 2 Moles of OH- ions

a) Before the addition of HCl, i.e.,when 0.00 mL of HCl has been added;

Concentration of hydroxide ions, [OH-] = 0.140 x 2 = 0.a) [OH-] = 0.100 x 2 = 0.280

pOH = -log [OH-]

pOH = -log (0.280)

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pH = 14 - 0.553= 13.447

b) pH when 15.0 mL HCl has been added

Moles Ba(OH)₂ = concentration × volume

Volume of Ba(OH)₂ = 82.0 mL = 0.082 L, concentration = 0.140 M

moles of Ba(OH)₂ = 0.082 x 0.140 = 0.01148 moles

moles OH- produced by 0.01148 moles of Ba(OH)₂ = 2 x 0.01148= 0.02296

moles HCl = 0.0150 L x 0.560 = 0.0084

moles of H+ produced by 0.0084 HCl = 0.0084

0.0084 H+ willnreact with 0.0084 moles of OH-

moles OH- left after the reaction = 0.02296 - 0.0084 = 0.01456 moles

total volume of new solution = (82 +15) mL = 97 mL => 0.097 L

Concentration of OH- ions = moles / volume

[OH-] = 0.01456 / 0.097 = 0.1501

pOH = -log [OH-]

pOH = -log (0.150)

pOH = 0.824

pH = 14 - 0.824

pH = 13.176

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