The strongly polar, hydrogen-bonding properties of water make it an excellent solvent for ionic (charged) species. By contrast,
nonionized, nonpolar organic molecules, such as benzene, are relatively insoluble in water. In principle, the aqueous solubility of any organic acid or base can be increased by converting the molecules to charged species. For example, the solubility of benzoic acid in water is low. The addition of sodium bicarbonate to a mixture of water and benzoic acid raises the pH and deprotonates the benzoic acid to form benzoate ion, which is quite soluble in water?
1 answer:
Answer:
hello your question is incomplete attached below is the complete question
Pyridinium ion is more soluble in 0.1 M HCL (aqueous solution) and this is because the Pyridine ion reacts with NaOH and after the reaction it comes out neutral.
N-acetyltyrosine methyl ester is more soluble in 0.1 M HCL while
B-napthol is more soluble in 0.1 M NaOH because it forms phenoxide ion
Explanation:
Pyridine ion is more soluble in 0.1 M HCL (aqueous solution) and this is because the Pyridine ion reacts with NaOH and after the reaction it comes out neutral.
N-acetyltyrosine methyl ester is more soluble in 0.1 M HCL while
B-napthol is more soluble in 0.1 M NaOH because it forms phenoxide ion
attached below is the remaining part of the solution
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Answer: a) 49.8 gram
b) 47.0 %
Explanation:
First we have to calculate the moles of glucose
The balanced chemical reaction will be,
From the balanced reaction, we conclude that
As,1 mole of glucose produce = 2 moles of ethanol
So, 0.54 moles of glucose will produce = mole of ethanol
Now we have to calculate the mass of ethanol produced
Now we have to calculate the percent yield of ethanol
Therefore, the percent yield is 47.0 %
Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.
In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)
pH = 2.88 ==>![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
= 
= 0.001 
The change in Concentration Δ![[CH_{3}COOH]](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D)
= 0.001
CH3COOH H+ CH3COOH
Initial 
0 0
Change -0.001 +0.001 +0.001
Equilibrium - 0.001 0.001 0.001
Since the
value is so small, the assumption
![[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D_%7Binitial%7D%20%3D%20%5BCH_%7B3%7DCOOH%5D_%7Bequilibrium%7D)
can be made.
![k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}](https://tex.z-dn.net/?f=%20k_%7Ba%7D%20%3D%20%5Btex%5D%3D%201.8%2A10%5E%7B-5%7D%20%20%3D%20%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%20%3D%20%20%5Cfrac%7B0.001%5E%7B2%7D%7D%7Bx%7D%20)
Solve for x to get the required concentration.
note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.
2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind
Hope this helps!
In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)
pH = 2.88 ==>
The change in Concentration Δ
CH3COOH H+ CH3COOH
Initial
Change
Equilibrium
Since the
Solve for x to get the required concentration.
note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.
2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind
Hope this helps!