
Use Le Châtelier’s principle to explain how the common ion effect affects the pH of a solution.

Interpretation:
By using Le Chatelier’s principle, the changes of pH of a weak acid solution due to common ion effect has to be explained.
Concept introduction:
- Le Chatelier's principle states that if a system in equilibrium gets disturbed due to modification of concentration, temperature, volume, and pressure, then it reset to counteract the effect of disturbance.
- When the concentration of one of the ions of a chemical solution got higher, it reacts with counter charged ions and precipitated out as salt till the ion product equals solubility product is called common ion effect.
- Incomplete dissociation of an acid in aqueous solution is called weak acid.
Answer to Problem 16.1QP
The pH of weak acid (acetic acid) increases due to addition of acetate ions from sodium acetate which suppress ionization of acetic acid leads to decrease in percent ionization of acetic acid. The equilibrium shifts towards left because of more acetate ion (common ion).
Explanation of Solution
To explain: The changes of pH of a weak acid solution due to common ion effect.
The equilibrium reaction of acetic acid and sodium acetate are as follows,
In an aqueous solution, dissolve acetic acid (weak acid) thoroughly which dissociates as acetate ions and hydronium ions. Then add sodium acetate (strong electrolyte) and it dissociates completely to form sodium ions and acetate ions. The common ion produced from both acetic acid and sodium acetate is acetate ion. By following the principle of Le Chatelier Principle, the acetate ions from sodium acetate combine with hydronium ions and the equilibrium shifts towards left which reduce the ionization of acetic acid. Thus lowers the percent dissociation of acetic acid and due to decrease in hydrogen ion concentration the pH of the solution will increase.
The change of pH of a weak acid solution due to common ion effect was explained by Le Chatelier's principle.
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Chapter 16 Solutions
CHEMISTRY (LL) W/CNCT >BI<
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- Decide whether these proposed Lewis structures are reasonable. proposed Lewis structure Is the proposed Lewis structure reasonable? Yes. :0: Cl C C1: 0=0: : 0 : : 0 : H C N No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* ☐ Yes. No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* Yes. ☐ No, it has the wrong number of valence electrons. The correct number is: ☐ No, it has the right number of valence electrons but doesn't satisfy the octet rule. The symbols of the problem atoms are:* | * If two or more atoms of the same element don't satisfy the octet rule, just enter the chemical symbol as many times as necessary. For example, if two oxygen atoms don't satisfy the octet rule, enter "0,0".arrow_forwardDraw the Lewis structure for the polyatomic trisulfide anion. Be sure to include all resonance structures that satisfy the octet rule. с [ ] - Garrow_forward1. Calculate the accurate monoisotopic mass (using all 1H, 12C, 14N, 160 and 35CI) for your product using the table in your lab manual. Don't include the Cl, since you should only have [M+H]*. Compare this to the value you see on the LC-MS printout. How much different are they? 2. There are four isotopic peaks for the [M+H]* ion at m/z 240, 241, 242 and 243. For one point of extra credit, explain what each of these is and why they are present. 3. There is a fragment ion at m/z 184. For one point of extra credit, identify this fragment and confirm by calculating the accurate monoisotopic mass. 4. The UV spectrum is also at the bottom of your printout. For one point of extra credit, look up the UV spectrum of bupropion on Google Images and compare to your spectrum. Do they match? Cite your source. 5. For most of you, there will be a second chromatographic peak whose m/z is 74 (to a round number). For one point of extra credit, see if you can identify this molecule as well and confirm by…arrow_forward
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