Volume of NaOH.docx

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RESULT 0 0.5 1 1.5 2 2.5 3 3.5 4 0 2 4 6 8 10 12 14 f(x) = 2.69380952380952 x + 3.84833333333333 R² = 0.865409334965303 Titration 1 Volume of NaOH (mL) pH pH= Equivalence point 1.72 ½ equivalen pH= 8.2 Volume of NaOH (mL) Titration 1 Titration 2 0 4.52 4.50 0.5 4.97 5.06 1.0 5.50 5.50 1.5 6.13 6.06 2.0 11.35 11.40 2.5 11.80 11.84 3.0 12.05 12.03 3.5 12.18 12.16

Transcript of Volume of NaOH.docx

Volume of NaOH (mL)Titration 1Titration 2

04.524.50

0.54.975.06

1.05.505.50

1.56.136.06

2.011.3511.40

2.511.8011.84

3.012.0512.03

3.512.1812.16

RESULT

Equivalence point equivalence pointpH= 8.21.72 mLpH= 8.4

8.424.21 equivalence point Equivalence point1.73 mL

Titration 1Titration 2

Volume of bases at equivalence point (mL)1.721.73

Volume of bases distance to equivalence point (mL)0.860.87

pH at equivalence point 8.48.42

pH at distance to equivalence point 4.24.21

Ka of acid6.31 x 10-56.17 x 10-5

DISCUSSION In this experiment, the acid ionization constants, Ka value of unknown acid solution is determined by using titration method of unknown acid with strong base which is 0.1 M sodium hydroxide, NaOH and by measuring the pH of the weak acid. Ka is a quantitative measure of the strength of an acid which indicates the relative strength of an acid. In titration method when strong base that is sodium hydroxide is added into a solution of weak acid, hydroxide ions, OH- from strong base will react with hydronium ions until it reached its equilibrium. The pH value will increase slowly and gradually when the number of moles of sodium hydroxide solution added is equals to the number of mole of unknown weak acid. Based on the experiment performed for the first titration, we can see that the pH changes very slowly and gradually. From the graph of the first titration we can see that there is an increment of pH value from 0 to 1.5 mL of NaOH but a sudden change of pH occur when 2.0 mL of NaOH is added and the pH increase from 6.13 to 11.35. From the graph, the equivalence point is determined which is 8.4. As we approach the equivalence point, the pH depends on the amount of weak acid present and the amount of conjugate base that is formed during the neutralization. At the equivalence point, all the weak acid is neutralized and converted to its conjugate base. The Ka of the unknown acid was calculated using the half-equivalence point which is 6.31 x 10-5. The nearest Ka value of the weak acid is benzoic acid which is 6.46x10-5Meanwhile for second titration, sudden change of pH occur when 2.0 mL of NaOH is added and the pH increase from 6.06 to 11.40. The equivalence point is 8.42 and from the equivalence point the Ka value calculated is 6.17 x 10-5. The nearest Ka value of the weak acid is benzoic acid which is 6.46x10-5

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