Vinegar Sample 1 pH = 3.211 2.89 mL Vinegar Sample 2 pH 3.102 3.00 mL pH = 4.895
ID: 1020195 • Letter: V
Question
Vinegar Sample 1 pH = 3.211 2.89 mL Vinegar Sample 2 pH 3.102 3.00 mL pH = 4.895 12.92 mL pH = 8.531 20.31 mlL inegar Sample 3 pH 3.105 1.61 mL pH = 4.687 9.50 mL pH = 8.934 Initial pH Reading Initial Buret Reading Midpoint pH Reading Midpoint Buret Reading Final pH Reading Final Buret Reading Volume of NaOH Consumed Calculated Molarity of Acetic Acid in the Vinegar Sample Volume NaOH Consumed at Midpoint Reading Moles of NaOH Consumed at Midpoint Reading [A- Moles of Acetic Acid Remaining at Midpoint HA Calculated pK pH 8.238 20.12 mlL 2 19.00 mLExplanation / Answer
If you have concentration of NaOH , then its midpoint can be found it easily.
Mid point is the point at which the half of the reactant is consumed .
As initial reading given as - Vi = 2.89 mL, Vf = 20.12 mL
The total volume added = 20.12 - 2.89 = 17.23 mL
in the above volume, exactly half of the volume should be consumed for mid point = 17.23 mL / 2 = 8.615 mL
Thus, Volume at mid point = 8.615 mL
So Burett reading at this point should be = 8.615 mL + initial reading = 8.615 mL + 2.89 mL = 11.50 mL
Moles of NaOH consumed = VNaOH (in L) * concentration (M) = "x" mols
Remaining acetic acid = Initial moles of acetic acid - Moles of NaOH consumed (x) = " y " mol
Since , we have half of the amount of acetic acid, pH can be calculated by using ICE table.
Therefore, pKa = pH (at mid point)
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To do above calculation, there must be concentration of NaOH. Please substitute the value of concentration in the above given information to get the answer.
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