. Consider a buffer made by mixing equal amounts of acetic acid and sodium aceta
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Question
. Consider a buffer made by mixing equal amounts of acetic acid and sodium acetate. Refer to equation 8-1 and describe the relative amounts of each of the three components present in the buffer solution, a. predict what will happen to the concentrations of each of the three components if a small amount of HCl is added, and b. predict what will happen if a small amount of NaOH is added. c. 2. Generate an outline of a procedure for preparing 100 mL of an acetic acid/acetate buf- 3. If a buffer is prepared using equal amounts of acetic acid and sodium acetate, predict 4. If a buffer is made up to be 1.0 M in acetic acid and 1.0 M in sodium acetate, predict fer what the pH of that solution will be. how much acid will be required to change the pH of 100 mL of the buffer by 0.1 pH units.Explanation / Answer
1a) The dissociation equations are given
HOAc -------> H+ + OAc-
NaOAc ---------> Na+ + OAc-
The pH of the buffer solution is given as
pH = pKa + log [NaOAc]/[HOAc] where pKa is the acid dissociation constant of HOAc
When equal amounts of HOAc and NaOAc are mixed (we will for simplicity, assume that equal amounts refer to equal number of moles of each component), the ratio of the molar concentration, i.e, [NaOAc]/[HOAc] is unity; this happens because both HOAc and NaOAc are dissolved in the same volume of the solution and hence the molar ratio is equal to the ratio of the moles. Consequently, we have log (1) = 0 and pH = pKa .
Since HOAc is a weak acid (poorly ionized) while NaOAc is a strong electrolyte (completely ionized), the solution will contain equal molar concentrations of HOAc, OAc- and Na+. Consequently, the relative amounts (in terms of moles) of each component will be equal.
b) Na+ is a spectator ion in the acetic acid/acetate buffer system and hence, its concentration and thus, the amount will remain the same as in part (a) above.
HCl is a strong electrolyte (completely ionized) and dissociates to furnish protons; these protons react with OAc- to give HOAc as below.
HCl -------> H+ + Cl-
H+ + OAc- -------> HOAc
Therefore, OAc- is neutralized by HCl and its concentration (or amount in moles) decreases while the concentration (or number of moles) of HOAc increases.
c) NaOH is a strong electrolyte and dissociates completely.
NaOH --------> Na+ + OH-
Upfront, the amount (number of moles) of Na+ is higher than that in part (a) above.
OH- picks up a proton from HOAc to form OAc-. Therefore, HOAc is neutralized and hence, its amount decreases. The reaction produces more OAc- and consequently, the amount of OAc- in the buffer increases.
OH- + HOAc -------> OAc- + H2O
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