Titration Worksheet Complete the data table below. Then plot the data points and

Question

Titration Worksheet Complete the data table below. Then plot the data points and draw a smooth curve connecting the points. Finally, label the equivalence point and half-equivalence point on your plot. The dissociation constant for acetic acid is, K, 1.8 x 10 Titration of 25 mL of 0.100 M acetic acid with 0.100 M sodium hydroxide 14 13 12 10 0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 Voume of NaOH added (mL) pH data for the titration of a weak acid (25 ml of 0.100M acetic acid) and a strong base (0.100 M NaOH). Volume of NaOH added (mL) 0.0 4.14 12.5 20.0 24.0 25.0 4.92 6.12 10.29 300 35.0 12.22 Chem 112-Spring 2014-Gard

Explanation / Answer

II will help you to complete the data, but the plot do it by yourself.

When no base is added the pH: (I will label acetic acid as HA)

r: HA <---------> A- + H+   Ka = 1.8x10-5 ---> pKa = 4.74 ----> Kb = 1x10-14/1.8x10-5 = 5.56x10-10 ---> pKb = 9.26
i: 0.1 0 0
e: 0.1-x x x

1.8x10-5 = x2/0.1-x
1.8x10-5 * 0.1 = x2
x = [H+] = 1.34x10-3 M
pH = -log(1.34x10-3) = 2.87

At the half equivalence point (12.5 mL) the pH is always the same as pKa. This is because the mole ratio between the reactant and the product is 1, because only the half has been consumed. pH = pKa = 4.74

At the equivalence point we only have the base in excess ans the reaction taking place is:
A- + H2O <------> HA + OH- Kb = 5.56x10-10

moles = 0.1 * 0.025 = 0.0025 moles
Concentration = 0.0025 / 0.050 L = 0.05 M

r: A- + H2O <------> HA + OH- Kb = 5.56x10-10
i: 0.05 0 0
e: 0.05-x x x

5.56x10-10 = x2 / 0.05
5.56x10-10 * 0.05 = x2
x = [OH] = 5.27x10-6 M
pOH = -log(5.27x10-6) = 5.28
pH = 14-5.28 = 8.72

Try to do the remaining point yourself with the explanation here, and plot the data. Hope this helps

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