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anyanavicka [17]
3 years ago
6

What is the molarity of a solution prepared by dissolving 54.3 g of Calcium nitrate

Chemistry
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer: 40 + 2x14 + 6x16 = 164g/mole

54.3g x [1mole / 164g] = 0.331moles

355mL x 1L / 1000mL = 0.355L

molarity = 0.331moles / 0.355L =

00

Explanation:

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To understand the relation between the strength of an acid or a base and its pKa and pKb values. The degree to which a weak acid
elena-14-01-66 [18.8K]

Answer:

pKa = 3.675

Explanation:

  • pKa = - Log Ka

∴ <em>C</em> X-281 = 0.079 M

∴ pH = 2.40

let X-281 a weak acid ( HA ):

∴ HA ↔ H+ + A-

⇒ Ka = [H+] * [A-] / [HA]

mass balance:

⇒<em> C</em> HA = 0.079 M = [HA] + [A-]

⇒ [HA] = 0.079 - [A-]

charge balance:

⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water

⇒ [H+] = [A-]

∴ pH = - log [H+] = 2.40

⇒ [H+] = 3.981 E-3 M

replacing in Ka:

⇒ Ka = [H+]² / ( 0.079 - [H+] )

⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )

⇒ Ka = 2.113 E-4

⇒ pKa = - Log ( 2.113 E-4 )

⇒ pKa = 3.675

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3 years ago
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