100.0 ml of 0.100 M acetic acid is placed in a flask and is titrated with 0.100

Question

100.0 ml of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 M sodium hydroxide. An appropriate indicator is used. Ka for acetic acid is 1.7 x 10 -5 Calculate the pH in the flask at the following points in the titration. a. when no NaOH has been added. b. after 25.0 ml of NaOH is added c. after 50.0 ml of NaOH is added d. after 75.0 ml of NaOH is added e. after 100.0 ml of NaOH is added f. what would be the appropriate indicator that was used? consult fig 16.7 in BLB and explain your choice. g. What would the pH be after 300 ml of 0.100 M NaOH was added? h. How could the equivalence point be detected without the use of a visual indicator?

Explanation / Answer

(a)

100 ml 0.1M

1.7 x 10 -5 =x2/(0.1)

x = 0.0013038 =conc of H+

So PH=-log(0.0013038)

PH=2.885

(b)

Initial moles of CH3COOH = volume x concentration

= 100/1000 x 0.100 = 0.01 mol

Moles of NaOH added = volume x concentration

= 50/1000 x 0.100 = 0.005 mol

CH3COOH + NaOH => CH3COOONa + H2O

Moles of CH3COOH left = initial moles of CH3COOH - moles of NaOH added

= 0.01 - 0.005 = 0.005 mol

Moles of CH3COONa formed = moles of NaOH added = 0.005 mol

Henderson-Hasselbalch equation:

pH = pKa + log([CH3COONa]/[CH3COOH])

= -log Ka + log(moles of CH3COONa/moles of CH3COOH)

= -log(1.7 x 10-5) + log(0.005/0.005)

= 4.77

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