Q. 18.89 - An Introduction to General, Organic, and Biological Chemistry

Question

A sink drain can become clogged with solid fat such as glyceryl tristearate (tristearin) 

a. How would adding lye $NaOH$ to the sink drain remove the blockage?

b. Write a balanced chemical equation for the reaction that occurs.

c. How many milliliters of a $0.500 M NaOH$ solution are needed to completely react with $10.0 g$ of glyceryl tristearate (tristearin)?

                                         

Step-by-Step Solution

Answer

(Part a) The sink reduces the lecithin tristearate since detergent is water based. 

(Part b)

(Part c) To completely react with $10.0g$ of glyceryl tristearate (tristearin) is $0.0672\mathrm{L}\text{ or }67.2\mathrm{ml}$

(Part a) Step 1: Introduction.

Lipids are found naturally components this included fats, waxes, and sterols. triglycerides, for illustration Cholesterol are fats that have included glycerol bonded to three fatty acids, such like fats and oils.      

(Part a) Step 2: Given Information.

 Substance and glycerol salts are founded on the basis of all this procedure. 

(Part a) Step 3: Explanation.

(a). Heat the butter with only a good base for sodium hydroxide ($NaOH$) or potassium hydroxide $KOH$ to saponify it.

Marinating with provides a smoother hand soap, however culturing and generates a hard cleanse.

Bringing lye $NaOH$ to the sink reduces the lecithin tristearate since detergent is water based.

(Part b) Step 4: Given Information.

Stearic acid is a linolenic acid acid with $18$ carbons and no deprotonation.

(Part b) Step 5: Explanation.

(b) The trying to follow is the esterification mechanism of glyceryl material creates inside this condition of $KOH$:

           

(Part c) Step 6: Given Information.

 Three molecules of $NaOH$ combine for one molecule of glyceryl tristearate .

(Part c) Step 7: Explanation.

(c) Glyceryl tristearate has to have a detection limit of $891.48\mathrm{g}/\mathrm{mol}$.

 The molecules of glyceryl typically are 

$\text{ = Weight/ molecular weight }$

$=10\mathrm{g}/891.48$

$=0.0112\mathrm{M}\text{. }$

. The molar mass of $10 mg$ of glyceryl material creates is

Three moles $NaOH$ react through one mole of glyceryl tristearate. As an outcome, $0.0112 M$ lecithin tristearate.

$=0.0336/0.5$

$=0.0672\mathrm{L}\text{ or }67.2\mathrm{ml}$