BRAINLIEST IF CORRECT
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The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).
The general reaction is:
2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)
We can write the above reaction in <u>two reactions</u>, one for oxidation and the other for reduction:
- Oxidation reaction
Li⁰(s) → Li⁺(aq) + e⁻ (2)
- Reduction reaction
Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)
We can see that Li⁰ is oxidizing to Li⁺ (by <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).
We must remember that the reducing agent is the one that will be oxidized by <u>reducing another element</u> and that the oxidizing agent is the one that will be reduced by <u>oxidizing another species</u>.
In reaction (1), the<em> reducing agent</em> is <em>Li</em> (it is oxidizing to Li⁺), and the <em>oxidizing agent </em>is<em> Fe(CH₃COO)₂</em> (it is reducing to Fe⁰).
Therefore, the reducing agent in reaction (1) is lithium (Li).
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Answer:- The Ka for the acid is .
Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:
HA(aq)\rightarrow H^+(aq) + A^-(aq)
Now, we make the ice table for this equation as:
HA(aq)\rightarrow H^+(aq) + A^-(aq)
I 0.25 0 0
C -X +X +X
E (0.25 - X) X X
where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.
X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.
Where, Ka is the acid ionization constant. Let's plug in the values.
Let's calculate the value of X first using the equation:
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on taking antilog ob above equation we get:
= 0.00195
So, X = 0.001195
Let's plug in this value of X in the equation:-
So, the value of Ka for butyric acid is .