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Andrews [41]
2 years ago
15

Give the expected organic product when 3-phenylpropanoic, PhCH2CH2COOH , is treated with NaOH, H20 .If more than one product is

possible, only draw the major product.Include counter-ions, e.g., Na+, I-, in your submission, but draw them in their own separate sketcher.Separate structures with + signs from the dropdown menu.

Chemistry
1 answer:
Ann [662]2 years ago
7 0

Answer:

when 3-phenylpropanoic, PhCH₂CH₂COOH , is treated with NaOH, H₂0 it produce sodium salt of 3 - phenylpropanoic acid with water as a major product.

Explanation:

Carboxylic acid when treated with a base like NaOH or KOH gives carboxylate ion because H⁺ from acid reacts with OH⁻ from base gives H₂0 and salt of sodium carboxylate.

The mechanism of the above reaction given below

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<h2>Question:- </h2>

A solution has a pH of 5.4, the determination of [H+].

<h2>Given :- </h2>
  1. pH:- 5.4
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<h2>To find :- concentration of H+</h2>

<h2>Answer:- Antilog(-5.4) or 4× 10-⁶</h2>

<h2>Explanation:- </h2><h3>Formula:- pH = -log H+ </h3>

Take negative to other side

-pH = log H+

multiple Antilog on both side

(Antilog and log cancel each other )

Antilog (-pH) = [ H+ ]

New Formula :- Antilog (-pH) = [+H]

Now put the values of pH in new formula

Antilog (-5.4) = [+H]

we can write -5.4 as (-6+0.6) just to solve Antilog

Antilog ( -6+0.6 ) = [+H]

Antilog (-6) × Antilog (0.6) = [+H]

Antilog (-6)  = {10}^{ - 6} ,  \\ Antilog (0.6)  = 4

put the value in equation

{10}^{ - 6}   \times 4 = [H+] \\ 4 \times   {10}^{ - 6}  = [H+]

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