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valentina_108 [34]
3 years ago
7

What is the molarity of a solution made by dissolving 8.60 g of a solid with a

Chemistry
1 answer:
Dima020 [189]3 years ago
8 0

Answer:

1.43 M

Explanation:

We'll begin by calculating the number of mole of the solid. This can be obtained as follow:

Mass of solid = 8.60 g

Molar mass of solid = 21.50 g/mol

Mole of solid =?

Mole = mass / molar mass

Mole of solid = 8.60 / 21.50

Mole of solid = 0.4 mole

Next, we shall convert 280 mL to litre (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

280 mL = 280 mL × 1 L / 1000 mL

280 mL = 0.28 L

Thus, 280 mL is equivalent to 0.28 L.

Finally, we shall determine the molarity of the solution. This can be obtained as illustrated below:

Mole of solid = 0.4 mole

Volume = 0.28 L

Molarity =?

Molarity = mole / Volume

Molarity = 0.4 / 0.28

Molarity = 1.43 M

Thus, the molarity of the solution is 1.43 M.

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2 years ago
If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a
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Answer:

pH = 4.543

Explanation:

  • CH3CH2COOH  + H2O ↔ CH3CH2COO-  +  H3O+
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∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ <em>C</em> CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ <em>C</em> CH3CH2COOH = 0.048 M

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mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543

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