Neutralization with equilibrium

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In summary, the question is how to calculate the pH after 10.0 mL of 0.4 M NaOH is added to 20.0 mL of 0.5 M CH3COOH. You should figure out how many millimoles of each reactant and how many milliliters of each reactant, do the stoichiometry to figure out how many mmol and mL there are of the conjugate base and acid, and then calculate the molarity of the conjugate base and the acid. Then, you should use the Henderson-Hasselbach equation to figure out the pH.
  • #1
dumbadum
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The question is:
calculate the pH after 10.0 mL of 0.4 M NaOH is added to 20.0 mL of 0.5 M CH3COOH (Ka CH3COOH = 1.8 x 10^-5)

I'm thinking that if I find the Kb of NaOH, then I can find how much of OH and H will be produced. I can then subtract the concentrations and find the pH or something. But how do I take into account the fact that CH3COONa is a basic salt and how to I find the Kb or NaOH?
Or in short, How would I go about the question?
Thank you!
 
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  • #2
Well, the reaction that will take place will be (I wrote the inorganic formula for acetic acid):

HC2H3O2 + OH- ---> C2H3O2- + H2O

Now, figure out how many millimoles of each reactant and how many milliliters of each reactant. Then, do the stoichiometry to figure out how many mmol and mL there are of the conjugate base and acid after the reaction goes to completion. Then, calculate the molarity of the conjugate base and the acid, and then you should be able to use the Henderson-Hasselbach equation to figure out the pH.

I hope this works =)
 
Last edited:
  • #3
goes like that

i guess it goes like that...
you have a solution 0,5 M of CH3COOH
this solution has 20 mL meaning u have 0,01 mol of the acid... when u mix it with the 10 mL from the NaOH the new solution will have a volume of 30 mL... so the new molarity of the acid is 0,33 (0,01 mol/0,03 L)
doing the same with the NaOH you find out u have 4.10^-3 mol, meaning the new solution will have a molarity of 0,13
the acid concentration is higher than the base´s; since they are together in the same recipient there will happen a neutralization reaction between them; since there is more acid than base, the final concentration of acid will be 0,01 - 4.10^-3 = 6.10^-3 moles of the acid in 30 mL..the final molarity will be 0,2
then u calculate it normaly
Ka = ([H+[ . [CH3COO-[) / [CH3COOH[
1,8.10^-5 . 0,2 = X^2
x = 1,897.10^-3
pH = 2,72
 
  • #4
(1) [tex] K_{b} [/tex] is an expression used only for weak bases. Since NaOH is a strong base (as it dissociates completely in water), it has no [tex] K_{b} [/tex] value.

(2) You should know that [tex] [H^+] = \frac{K_{a}[CH_{3}COOH]}{[CH_{3}COO^-]} [/tex]. So, can you figure out [tex] [CH_{3}COO^-]\ and\ [CH_{3}COOH] [/tex] once the neutralisation reaction is complete?

(3) You are right in saying that [tex] CH_{3}COONa [/tex] is a basic salt. However, [tex] CH_{3}COO^-\ and\ CH_{3}COOH [/tex] exist in equilibrium, and the [tex] K_{a} [/tex] formula has already taken into account the basicity of [tex] CH_{3}COO^- [/tex] when calculating the pH, so one formula is all you need!

HINT: NaOH is the limiting reagent in the neutralisation reaction.

NOTE: When calculating the final concentrations, remember that the total volume of the solution has changed.

POSTSCRIPT: (Further explanation for comment 3) Observe the [tex]K_{a}[/tex] formula above. Note that if [tex] [CH_{3}COO^-] [/tex] increases, [tex] [H^+] [/tex] will decease and the pH will increase. The reverse is also true. This is why I say that "the [tex] K_{a} [/tex] formula has already taken into account the basicity of [tex] CH_{3}COO^- [/tex]"

Hope this helps!
 
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What is neutralization with equilibrium?

Neutralization with equilibrium is a chemical process in which an acid and a base react to form a salt and water. This reaction results in the formation of equal amounts of hydrogen ions (H+) and hydroxide ions (OH-) in solution, leading to a neutral pH.

How does neutralization with equilibrium occur?

Neutralization with equilibrium occurs when an acid and a base are mixed together in equal proportions. The acid donates a proton (H+) to the base, forming water, while the base accepts the proton to form a salt. The reaction continues until there are equal amounts of H+ and OH- ions, resulting in a neutral solution.

What is the importance of neutralization with equilibrium?

Neutralization with equilibrium is important in many aspects, including industrial processes, medical treatments, and environmental applications. In industries, it is used to neutralize harmful acids and bases, while in medicine, it is used to treat acid-related conditions such as heartburn. It also helps maintain the pH balance in natural water bodies.

What factors affect neutralization with equilibrium?

The rate of neutralization with equilibrium can be affected by several factors, including temperature, concentration of reactants, and presence of catalysts. Higher temperatures increase the rate of reaction, while higher concentrations of reactants lead to a faster reaction. Catalysts can also speed up the reaction by reducing the activation energy required.

How is neutralization with equilibrium different from titration?

Neutralization with equilibrium is a specific type of chemical reaction between an acid and a base, resulting in a neutral pH. On the other hand, titration is a technique used to determine the concentration of an unknown solution by reacting it with a known solution. While both involve acid-base reactions, titration is a quantitative method, whereas neutralization with equilibrium is a qualitative process.

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