Determination of the Ionization Constant for a Weak Acid from a Titration Curve

Question

Determination of the Ionization Constant for a Weak Acid from a Titration Curve

fill out data table based on data:

1st derivative curve and data:

Data table:

NaOH mL pH 0 2.71 0.5 2.92 1.5 3.19 2 3.48 2.5 3.58 3 3.66 3.5 3.73 4 3.81 4.5 3.88 5 3.95 5.5 3.99 6 4.04 6.5 4.08 7 4.13 7.5 4.18 8 4.22 8.5 4.24 9 4.26 9.5 4.34 10 4.38 10.5 4.41 11 4.44 11.5 4.47 12 4.52 12.5 4.55 13 4.59 13.5 4.61 14 4.65 14.5 4.69 15 4.72 15.5 4.75 16 4.81 16.5 4.84 17 4.87 17.5 4.9 18 4.93 18.5 4.98 19 5.02 19.5 5.07 20 5.1 20.5 5.16 21 5.2 21.5 5.26 22 5.33 22.5 5.45 23 5.46 23.5 5.54 24 5.63 24.5 5.71 25 5.86 25.5 6.07 26 6.3 26.5 7.09 27 10.41 27.5 11.17 28 11.5 28.5 11.68 29 11.82 29.5 11.87 30 11.96 30.5 12.01 31.5 12.09 32.5 12.17 33.5 12.23 34.5 12.27 35.5 12.31

Explanation / Answer

molarity of std NaOH ( N1)= 0.1N

volume of base to reach eq. pt (V1)=36.2ml

molarity of unknown acid V1N1 = V2N2

molarity of std Unknown acid ( N2)= ?

volume of acid taken (V2)=20ml

  ( N2)= V1N1 / V2

= 0.1x36.2 / 20

= 0.181N

MOLARITY OF UNKNOWN ACID = 0.181N, If molarity of NaOH is 0.1N and volume of unknown acid taken = 20ml

PKa = PH + log [acid] / [base]

[acid] = concentration of acid

[base] = concentration of base

PKa of unknown acid = 8(from first graph) +log 0.181/0.1

= 8+0.2576 = 8.2576

PKa = -log (Ka)

Ka of unknown acid = antilog (- PKa)

= antilog(-8.2576)

Ka of unknown acid = 5.5258 x 10-9

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