If carbon dioxide was left in the cola then the NaOH would have reacted with

If carbon dioxide was left in the cola then the naoh

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If carbon dioxide was left in the cola, then the NaOH would have reacted with the gas based upon the following equation: CO 2 (g) + NaOH (aq) NaCO 3 + H 2 O. The premise of this lab was based upon the idea that the phosphoric acid was the only compound in the cola that was being titrated, so the moles of NaOH used for the titration could be used to determine the moles of phosphoric acid present. If the NaOH could have possibly reacted with the carbonic acid as well, then the titration would have been inconclusive since there would have been no way to tell whether the carbon dioxide or phosphoric acid was reacting with the strong base.
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Calculations: Standardized Molarity of “0.01 M” NaOH solution (trial 1) NaOH (aq) + HKC 8 H 4 O 4 (s) → H 2 O (l) + KC 8 H 4 O 4 + (aq) Mass HKC 8 H 4 O 4 = 0.063 grams Moles HKC 8 H 4 O 4 = 0.063 1 mole 204.22 g = 0.000308 moles HKC 8 H 4 O 4 Moles NaOH = 0.000308 moles HKC 8 H 4 O 4 × 1 mol NaOH 1 mole HK C 8 H 4 O 4 = 0.000308 moles NaOH Concentration NaOH = 0.000308 moels NaOH 0.031 LNaOH = 0.010 moles/liter NaOH Initial Concentration of Phosphoric Acid in Cola Sample (graphically) At the first equivalence point, the moles of NaOH added is equal to the moles of phosphoric acid. First equivalence point occurred where the derivative function reached a relative maximum. This occurred when 8.9 mL of NaOH were added to the solution. 0.0089 L X 0.01 moles liter NaOH = 0.000089 moles phosphoric acid The volume of the phosphoric acid was 20 mL Concentration phosphoric acid = 0.00089 moles phosphoric acid 0.020 LCola = 4.5 X 10 -3 moles liter phosphoric acid Initial Concentration of Phosphoric Acid in Cola Sample (non graphically) K a1 = + ¿ H 3 O ¿ ¿ ¿ H 2 PO 4 ¿ ¿ H [ ¿¿ 3 PO 4 ] ¿ ¿ = H [ ¿¿ 3 PO 4 ] initial x [ x ] [ x ] ¿ H [ ¿¿ 3 PO 4 ] ¿ = 4.5 X 10 -3 moles liter phosphoric acid + ¿ H 3 O ¿ ¿ = x = 10 -2.4 = 0.00380 K a1 = [ 0.00380 ] [ 0.00380 ] 0.0045 0.00380 K a1 = 2.1 X 10 -2
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Percent Dissociation of H 3 PO 4 Percent dissociation = + ¿ H 3 O ¿ ¿ ¿ ¿ X 100 = + ¿ H 3 O ¿ ¿ = 10 -pH , pH = 2.42 + ¿ H 3 O ¿ ¿ = 10 -2.42 = 0.00380 moles liter [ H 3 PO 4 ] = 4.5 X 10 -3 moles liter Percent dissociation = 0.00380 0.0045 X 100 = 84%
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References: Lab Manual Bottomley, L.; Bottomley, L.A.; CHEM 1211K/1212K, Chemical Principles I & II Laboratory Manual, 2013-2014, p. 189-196. pK a and K a Values Ka and PKa for Polyprotic Acids." Upper Canada District School Board, n.d. Web. <;.
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  • Fall '12
  • SOPER
  • pH, Phosphoric acid

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