Question

In: Chemistry

For an acidic component, after treatment of a strong base the solution is then treated with...

For an acidic component, after treatment of a strong base the solution is then treated with a strong acid. Why is this step perfromed?

Solutions

Expert Solution

For an acidic component, first we have to neutralize with the strong base saturated NaOH or saturated KOH solution. In this case neutralization takes place in which, salts and water are produced. In such condition that acid component will be in water (neutral pH = 7) layer. So, it cannot be extracted further because the component will go into the water. To resolve this problem further we need to add strong acid so that it can come in acidic region (pH = 2 to pH = 3) from neutral solution region (pH = 7). Now your component is in acidic medium. After adding the organic solvent like ethyl acetate, chloroform etc. the acidic component will come into the organic layer. So, now you can easily extract the organic layer from the aqueous layer by washing with ethyl acetate 2 to 3 times. And finally add the Na2SO4 (to remove the excess water from the organic layer) into that organic layer and evaporate it. Final compound will be desired component.


Related Solutions

pH of the Buffer after the addition of strong base Original buffer solution= 50.0 mL of...
pH of the Buffer after the addition of strong base Original buffer solution= 50.0 mL of 1.0 M CH3COOH 50.0 mL of 1.0 M NaCH3COO Take 50.0 mL of this buffer solution and add 2.0 mL of 1.0 M NaOH Calculate the expected pH of this solution.
Explain why the addition of a small amount of base to an acidic buffer solution causes...
Explain why the addition of a small amount of base to an acidic buffer solution causes only a small change in the pH of the buffer solution. You should use any appropriate chemical and/or mathematical equations to substantiate your arguments
What is the pH of a weak acid strong base titration after each of the following...
What is the pH of a weak acid strong base titration after each of the following additions? Ka for HF is 3.5 x 10-4. a. 10.0 mL of 0.250 M NaOH added to 15.0 mL of 0.400 M HF b. 25.0 mL of 0.250 M NaOH added to 35.0 mL of 0.1785 M HF c. 20.0 mL of 0.250 M NaOH added to 20.0 mL of 0.350 M HF Please show your work so I can better understand this. thanks!...
For each strong base solution, determine [OH−] , [H3O+] , pH , and pOH . Part...
For each strong base solution, determine [OH−] , [H3O+] , pH , and pOH . Part A 0.40 M NaOH , determine [OH−] and [H3O+] . Express your answers using two significant figures. Enter your answers numerically separated by a comma. Part B For this solution determine pH and pOH . Express your answers using two decimal places. Enter your answers numerically separated by a comma. Part C 2.0×10−3 M Ca(OH)2 , determine [OH−] and [H3O+] . Express your answers...
For each strong base solution, determine [OH−] , [H3O+] , pH , and pOH . Part...
For each strong base solution, determine [OH−] , [H3O+] , pH , and pOH . Part A 8.77×10−3 M LiOH , determine [OH−] and [H3O+] . Express your answers using three significant figures. Enter your answers numerically separated by a comma. Part B For this solution determine pH and pOH . Express your answers using three decimal places. Enter your answers numerically separated by a comma. Part C 0.0312 M Ba(OH)2 , determine [OH−] and [H3O+] . Express your answers...
Acidic solution In acidic solution, the iodate ion can be used to react with a number...
Acidic solution In acidic solution, the iodate ion can be used to react with a number of metal ions. One such reaction is IO3−(aq)+Sn2+(aq)→I−(aq)+Sn4+(aq) Since this reaction takes place in acidic solution, H2O(l) and H+(aq) will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation: IO3−(aq)+Sn2+(aq)+ −−−→I−(aq)+Sn4+(aq)+ −−− Part A What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l)...
classify each substance as a strong acid , weak acid , strong base or weak base...
classify each substance as a strong acid , weak acid , strong base or weak base KOH, Ca(OH)2, H 2SO4, HI, HCOOH, NaOH, HCN, C5H5N
Classify each substance as a strong acid , weak acid strong base or weak base for...
Classify each substance as a strong acid , weak acid strong base or weak base for the following compounds HI C5H5N NaOO HCOOH KOH Ca(OH)2 H2SO4 HCN
Buffer Solution from Strong Base and Weak Acid Solution= 30.0 mL of 1.0 M CH3COOH 5.0...
Buffer Solution from Strong Base and Weak Acid Solution= 30.0 mL of 1.0 M CH3COOH 5.0 mL of 1.0 M NaOH 1. Calculate the total volume. 35 mL 2. What is the concentration of acetic acid in this solution after mixing? (include units) 3. What is the concentration of sodium hydroxide in this solution after mixing (include units) 4. Calculate the expected pH of this buffer solution. 5. Measured pH of this buffer solution. 3.89
A solution contains 0.1252 grams of strong base calcium hydroxide completely dissolved in 1255 ml of...
A solution contains 0.1252 grams of strong base calcium hydroxide completely dissolved in 1255 ml of pure water. What is the pH of this solution?
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT