Could someone please help me with this chemistry problem? Thank you! After perfo

Question

Could someone please help me with this chemistry problem? Thank you! After performing a titration using 20 mL of unknown weak acid with 0.3830M NaOH, I obtained the following results:

At Equivalence Point: Volume of NaOH = 4.835 mL;

At the Halfway Point; Volume of NaOH = 2.418 mL:

Ka = 7.08 x 10^-6 Initial [HA] = 0.0926M (determined by M1V1=M2V2 at the equivalence point)

Molar Mass = 1079.8 g/mol (determined by using original mass 1.9977 g/1.85 x 10^-3 moles -----------------------------------------------------------------------------------------------------------------------------------------------

Now I am asked the following questions in order to check my results for consistency; Using your experimental Ka value, the concentration of your unknown acid solution and the concentration of the standardized NaOH solution, CALCULATE the pH of the following points on your titration curve:

1) 120% of the way to the equivalence point

2) At the equivalence point

3) at the beginning of the titration, before any NaOH was added

These values are supposed to be calculated - not given by pH data collected. Thank you!

Explanation / Answer

1) 120% of the way to the equivalence point

if 20% excess, then

mmol of acid = MV = 0.0926*20 = 1.852 mmol of acid

mmol of base = 1.852 mmo lfor 100%

then 120% --> 120/100*1.852 = 2.2224 mmol of base required

mmol of OH- left = 2.2224-1.852 = 0.3704

[OH-] = mmol of OH- / Vtotal = 0.3704 / Vtotal

Vtotal = Vacid + Vbase = 20 mL + Vbase

Vbase = mmol of base/ Mbase * 1.2 = 2.2224 /0.383 = 5.802 mL --> 120% excess = 5.802 *1.2 = 6.9624 mL

[OH-] = mmol of OH- / Vtotal = 0.3704 / (6.9624+20) = 0.01373 M

pH = 14 + log[OH]

pH = 14 + log(0.01373) = 12.137

2) At the equivalence point

Vtoal = Vacid + Vbase = 5.802 + 20 = 25.802 mL

[A-] = 1.852 /25.802 = 0.0717

A- + H2O <->HA + OH-

Kb = [HA][OH-]/[A-]

(10^-14)/(7.08*10^-6) = x*x/(0.0717-x)

x = 1.006*10^-5

pOH = -log(1.006*10^-5) =

pH = 14-4.997 = 9.003

3) at the beginning of the titration, before any NaOH was added

initially, there is only acid

HA <->H+ + A-

Ka = [H+][A-]/[HA]

[HA] = mmol/V = 1.852 /20 =

7.08*10^-6 = x*x/(0.0926-x)

x = 8.06*10^-4

pH = -log( 8.06*10^-4) = 3.09

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