Volume of HCl at equivalence point: 18 ml -Show the calculations that are used t
ID: 505935 • Letter: V
Question
Volume of HCl at equivalence point: 18 ml
-Show the calculations that are used to experimentally determine from the titration data the
acetate concentration of the buffer:
-How does the experimentally determined acetate concentration compare to the assigned concentration?
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Volume of NaOH at equivalence point: 19 ml
-Show the calculations that are used to determine the acetic acid concentration of the buffer from the titration data:
-How does the experimentally determined acetic acid concentration compare to the assigned concentration?
9:22 PM Goal5.xlsx Done HC NaOH Sheet3 NaOl (M) 0.125 0,025 Orig Butler Vol 0.0000175 moles A- 0.0025 moles HA 0,0025 4.76 Vol HOT Added (mL) moles HCI added Cl limiting? pH Theoretical pH Actual 0.00E+00 Yes 4,66 1.000 25E-04 Yes 2.00 2.50E-04 Yes 4.75 3.75E-04 Yes 4.8 5.00E-04 Yes 4.00 4,84 5.000 6.25E-04 Yes 4.89 7.50E-04 Yes 4.94 7.00 8.75E-24 Yes 1.00E-03 Yes 5.06 8.00 5.12 9.00 1.12E-03 Yes 1.25E-03 Yes 5.19 1.38E-03 Yes 5.26 50F-03 Yes 5.34 1.62E-03 Yes 3.000 5.42 4.00 1.75E-03 Yes 5.19 5.00 5.6 1.88E-03 Yes 2.00E-03 Yes 5.79 17.00 2.12E-03 Yes 6.02 8.00 2.25F-03 Yes 6.48 10.86 2,38E-03 Yes 2.44E-03 Yes 9.50 20.00 2.50E-03 No 20.50 2.56E-03 No 2.62E-03 No 2.69E-03 No 22.00 2.75E-03 No N3 2.81E-03 No 87 23.00 2.88E-03 No 23.50 2.94E-03 No 24.00 3.00E-03 No 306E-03 No 2.02 3.12E-03 No 12.05 3.19E-03 No 25.50 2.08 3.25E-03 No Meme O O OExplanation / Answer
The HCl titration is used to calculate the mass of acetate anion (A-) in the buffer as per the reaction
A- + HCl -----> HA + Cl-
The equivalence point is the point on the curve where the pH changes sharply. We note that point to be the midpoint of 18 mL and 18.5 mL of HCl when the pH changes to 1.46 from 2.91. Therefore, the equivalence point is 18.25 mL.
Concentration of HCl = 0.135 M.
Moles of HCl consumed at the equivalence point = (18.25 mL)*(1 L/1000 mL)*(0.135 mol/L) = 0.00246375 mol.
At the equivalence point, we have
Moles of HCl consumed = moles of acetate present.
Therefore, moles of acetate present = 0.00246375 mol.
Concentration of acetate = moles of acetate/(volume of buffer) = (0.00246375 mol)/(0.025 L) = 0.09855 mol/L = 0.09855 M (ans).
The desired concentration of acetate is 0.10 M.
The percent error between the desired and actual concentration is (0.100 M – 0.09855 M)/(0.10 M)*100 = 1.45% (ans).
The NaOH titration is used to calculate the mass of acetic acid (HA) in the buffer as per the reaction
HA + NaOH -----> NaA + H2O
The equivalence point is the point on the curve where the pH changes sharply. We note that point to be the midpoint of 18 mL and 19 mL of NaOH when the pH changes to 6.48 from 10.86. Therefore, the equivalence point is 18.50 mL.
Concentration of NaOH = 0.125 M.
Moles of NaOH consumed at the equivalence point = (18.50 mL)*(1 L/1000 mL)*(0.125 mol/L) = 0.0023125 mol.
At the equivalence point, we have
Moles of NaOH consumed = moles of acetic acid present.
Therefore, moles of acetic acid present = 0.0023125 mol.
Concentration of acetic acid = moles of acetic/(volume of buffer) = (0.0023125 mol)/(0.025 L) = 0.0925 mol/L = 0.0925 M (ans).
The desired concentration of acetic is 0.10 M.
The percent error between the desired and actual concentration is (0.100 M – 0.0925 M)/(0.10 M)*100 = 7.50% (ans).
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