From the pH, calculate the concentration of the hydrogen ion. Next find the conc
ID: 497350 • Letter: F
Question
From the pH, calculate the concentration of the hydrogen ion. Next find the concentration of the acetate ion. Next determine the actual concentration of the acetic acid that is in the solution.
4. Using the following titration curve of a 25.00 mL sample of 0.1000 M acetic acid titrated with a 0.1000 M sodium hydroxide solution, label the four regions of the titration curve and fill in the table below. 14 Volume of NaOH added (mL) Moles of Identify the limiting Titration Volume Moles of pH reactant and the non- Curve of titrant sodium acetic acid hydroxide remaining limiting reactant used Region mL added (mo mol LR: Initial NonLR: LR: Buffer Non LR: LR: Equivalence point Non LR: LR: Post- Equivalence NonLR: pointExplanation / Answer
Initially:
V = 0 mL,
mol of NaOH = MV = 0.1*0 = 0
mol of acid = MV = 25*0.1 = 2.5*10^-3 mol of acid
pH = from:
Ka = [H+][A-]/[HA]
1.8*10^-5 = x*x/(0.1-x)
x = [H+] = 0.00133; pH = -log(0.00133) = 2.8761
limiting reactant --> base, since it is all cnsumed, therefore acid is excess
BUFFER:
0 < V < 25 mL will do... for simplicity... let us use 12.5 mL
so
mol of acid = MV = 25*0.1 = 2.5*10^-3 mol
mol of base = MV = 12.5*0.1 = 1.25*10^-3 mol
after reaction, there is:
weak acid = 1.25*10^-3
conjugate base 1.25*10^-3
pH = pKa + log(A-/HA)
pKa = 4.75 for acetic acid s
pH = 4.75 + log( 1.25*10^-3 / 1.25*10^-3)
pH = 4.75
limiting reactant is still base, since there is acid left
In equivalence point
there is no limiting reactant since there is stoichiometric ratio
V = 25 mL from the graph
mol of acid = MV = 25*0.1 = 2.5*10^-3 mol
mol of base = MV = 25*0.1 = 2.5*10^-3 mol
pH = from ionization:
A- + H"O <-> HA + OH- Kb
Kb = 5.55*10^-10 for acetic acid
Kb = [HA][OH-]/[A-]
5.55*10^-10 = x*x/(0.1/2 -x)
x = OH 5.26*10^-6
pOH = -log( 5.26*10^-6) = 5.279
pH = 14-5.279= 8.721
post equivalence point:
excess is base, then limiting is acid
Volume is V > 25 mL valid
mol of acid = MV = 25*0.1 = 2.5*10^-3 mol
mol of base = MV varies
pH ... > 8.7
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