pate: Lab Section Name Partner\'s name 1. Use your and data table to co the volu
ID: 511833 • Letter: P
Question
pate: Lab Section Name Partner's name 1. Use your and data table to co the volume ofNaoHtitrant you recorded before and after the large increase in the addition of 0.5 mL ofNaoH solution. (Remember to include your graphs with the report sheets) 2. Determine the volume of NaoH added at each equivalence point. To do this, add the two values determined above and divide by two. 3. Calculate the number of moles ofNaoH used. 4. See the equation for the neutralization reaction given in the introduction. Determine the number of moles off H3PO4 reacted. 5. Recall that you pipeted out 40.0 mLofthe beverage for the titration. Calculate the HalPO4 concentration. 6. pH PKa at the half equivalence point. From pKa you can determine Ka Trial 2 Trial 1 First Equivalence point 050 050M Concentration of NaOH (M) LlomL Volume of Cola (mL) NaOH volume added immediately before the large pH increase for the equivalent (mL) NaOH volume added immediately after the large pH uivalent (mL increase for the first volume of NaoH at the first equivalence point (mL) L amb Moles of NaOH at first equivalence point (mol) Moles of Phosphoric acid from first equivalence point volume of NaoH at equivalence point (mL) 2 mL 3,5 m L 3.2 pH (pK 1) at h first equivalence point Concentration of Phosphoric acid in Cola (M) first equivalenceExplanation / Answer
determination of H3PO4 in cola by NaOH is acid- base reaction in the ration of 3:1 as below
H3PO4+ 3NaOH ------------>Na3PO4 + 3H2O
calculation : no of moles = ( Molarity * vol. in Litres )
for NaOH (No. of moles at equivalence point) = 0.05 M * 0.004 L = 2.0 X 10-4 for trial 1 volume (A)
= 0.0.5 M * 0.007 L =3.5 X 10-4 for trial 2 volume (B)
for H3PO4( no. of moles ) = no. of moles of NaOH / 3 Because this is 3:1 reaction
= A / 3 = C for trial 1 volume
= B / 3 = D for trial 2 volume
Likewise have to calculate for second equivalence point as above
calculation 2: pH = -log(H+) but at 1/2 equivalence point pH =pKa so
pH=pKa=-log(ka1)=3.62
ka1= 2.4 X 10-3 trial 1 & 5 X 10-5 for trial 2
repeat the same calculation for second table
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