From the pH, calculate the concentration of the hydrogen ion. Next find the conc

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: point

Explanation / 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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