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wariber [46]
3 years ago
14

The weak acid HA is 2% ionized (dissociated) in a 0.20 M solution.

Chemistry
1 answer:
nekit [7.7K]3 years ago
7 0

Answer:

1. Ka = 8.16x10⁻⁵

2. pH = 2.40

Explanation:

1. The dissociation of a weak acid in water occurs as follows:

HA ⇄ H⁺ + A⁻

Ka = [H⁺] [A⁻] / [HA]

As 2% of the 0.20M solution is dissociated:

[H⁺] = [A⁻] = 0.20M * 2% = 0.004M -As H⁺ and A⁻ comes from the same reaction, their concentrations are the same

[HA] = 0.20M * 98% = 0.196M

Ka = (0.004)² / (0.196M) = 8.16x10⁻⁵

2. pH = -log [H⁺] = -log [0.004M]

<h3>pH = 2.40</h3>
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For the neutralization reaction involving hno3 and ca(oh)2, how many liters of 1. 55 m hno3 are needed to react with 45. 8 ml of
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For the neutralization reaction involving HNO₃ and Ca(OH)₂, liters of 1.55 M HNO₃ are needed to react with 45.8 ml of a 4.66 m Ca(OH)₂ solution is 0.270 L.

<h3>What is neutralization reaction?</h3>

Those reactions in which acids and bases will combine with each other for the formation of water molecule and salt, is known as neutralization reaction.

Given chemical reaction is:

2HNO₃ + Ca(OH)₂ → 2H₂O + Ca(NO₃)₂

Moles (n) of Ca(OH)₂ will be calculated by using the below formula:
M = n/V, where

M = molarity = 4.66M

V = volume = 45.8 mL = 0.045 L

n = (4.66)(0.045) = 0.2097 mol

From the stoichiometry of the reaction,

1 mole of Ca(OH)₂ = reacts with 2 moles of HNO₃

0.2097 mole of Ca(OH)₂ = reacts with 2×0.2097=0.419 moles of HNO₃

Again by using the molarity equation volume will be calculated for HNO₃ as:

V = (0.419) / (1.55) = 0.270 L

Hence required volume of HNO₃ is 0.270 L.

To know more about neutralization reaction, visit the below link:

brainly.com/question/23008798

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