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Vlad [161]
3 years ago
15

Using the 5M NaCI and 10% glucose stock solutions you generated, how would you prepare 100 ml of a solution that is both 150mM N

aCl and 1 % glucose?
Chemistry
1 answer:
fredd [130]3 years ago
4 0

Answer:

Take 3 mL of the 5 M NaCl solution, 10 mL of the 10% glucose solution, and add water for a final volume of 100 mL.

Explanation:

  • In order to calculate the required volume of the 5 M NaCl solution, we calculated the moles contained in a 100 mL solution that has a concentration of 150 mM:

0.1 L * 0.150 M = 0.015 moles of NaCl

With those moles we can calculated the required volume, using the concentration of the stock solution:

0.015 mol / 5 M = 0.003 L = 3 mL.

  • To make a solution that has a 1 % concentration of glucose, from a 10 % glucose solution, is the same as to make it ten times less concentrated. Thus, with a final volume of 100 mL, you would need to take 10 mL of the 10% glucose solution, because 100mL * 10/100 = 10.

So in order to prepare the solution, you would need to take 3 mL of the 5 M NaCl solution, 10 mL of the 10% glucose solution, and add water for a final volume of 100 mL.

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Answer:

a) C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

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Explanation:

a) balanced equation:

C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

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⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)

b) (2) in (1):

⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]

⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]

⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5

⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0

⇒ [ H3O+ ] = 2.517 E-3 M

c) pH = - log [ H3O+ ]

⇒ pH = - Log ( 2.517 E-3 )

⇒ pH = 2.599

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