You are to assume that the buffer is made up by mixing volumes of 0.100 M acetic
ID: 493704 • Letter: Y
Question
You are to assume that the buffer is made up by mixing volumes of 0.100 M acetic acid and 0.100 M sodium acetate solutions.
Calculate the volume of 0.100 M acetic acid required to prepare 60.0 mL of a buffer of pH 5.0.
ANSWER=21.4 mL
Calculate the volume of 0.100 M sodium acetate required to prepare 60.0 mL of a buffer of pH 5.0.
ANSWER=38.6 mL
Calculate the concentration [HA] in this buffer.
ANSWER=0.0357 M
Calculate the concentration [A-] in this buffer.
ANSWER=0.0643 M
Calculate the volume of 0.1000 M NaOH which would be required to titrate 20.0 mL of the buffer solution to the Basic equivalence point.
ANSWER=7.1 mL
THIS IS WHAT I DONT KNOW
Calculate the pH at the Basic equivalence point
Calculate the volume of 0.1000 M HCl which would be required to titrate 20.0 mL of the buffer solution to the Acidic equivalence point.
ANSWER=12.9 mL
ALSO THIS QUESTION
Calculate the pH at the Acidic equivalence point.
Explanation / Answer
Moles of HA in 20 mL of the buffer solution = (molar concentration of HA)*(volume of buffer solution in L) = (0.0357 mol/L)*(20 mL)*(1 L/1000 mL) = 0.000714 mole.
Moles of A- in 20 mL of the buffer solution = (molar concentration of A-)*(volume of buffer solution in L) = (0.0643 mol/L)*(20 mL)*(1 L/1000 mL) = 0.001286 mole.
The basic neutralization reaction is
HA + NaOH -----> NaA + H2O
Moles of NaOH added = moles of HA completely neutralized = 0.000714 mole.
Molar concentration of NaOH = 0.1000 M.
Therefore, volume of NaOH required = (moles of HA neutralized)/(molar concentration of NaOH) = (0.000714 mole)/(0.1000 mol/L) = 0.00714 L = (0.00714 L)*(1000 mL/1 L) = 7.14 mL (ans).
At the basic neutralization point, we have 0.000714 mole of HA completely neutralized to form 0.000714 mole of A-. Therefore, the total number of moles of A- at the basic neutralization point is (0.000714 + 0.001286) = 0.0020 mole.
Total volume of the solution = (20 + 7.14) mL = 27.14 mL.
Molar concentration of A- at the basic equivalence point = (moles of A-/total volume of the solution) = (0.0020 mole)/(27.14 mL) = 7.369*10-5 mM.
A- is the conjugate base of HA and establishes equilibrium as
A- + H2O ------> HA + OH-
Since OH- is formed, we need to work with Kb of A-. Given Ka = 1.76*10-5, we have
Kb = Kw/Ka = (1.0*10-14)/(1.76*10-5) = 5.68*10-10
Therefore, Kb = [HA][OH-]/[A-] = (x).(x)/(7.369*10-5 – x)
Assume x is small (since Kb is small) and hence
5.68*10-10 = x2/(7.369*10-5)
===> x2 = 4.18559*10-14
===> x = 2.04587*10-7
Therefore, [OH-] = 2.04587*10-7 and pOH = -log [OH-] = -log (2.04587*10-7) = 6.689
Thus, pH = 14 – pOH = 14 – 6.689 = 7.311 7.3 (ans).
The acid neutralization reaction is
A- + HCl -----> HA + Cl-
Moles of HCl added = moles of A- completely neutralized = 0.001286 mole.
Molar concentration of HCl = 0.1000 M.
Therefore, volume of HCl required = (moles of A- neutralized)/(molar concentration of HCl) = (0.001286 mole)/(0.1000 mol/L) = 0.001286 L = (0.001286 L)*(1000 mL/1 L) = 12.86 mL 12.9 mL (ans).
At the acid neutralization point, we have 0.001286 mole of A- completely neutralized to form 0.001286 mole of HA. Therefore, the total number of moles of HA at the acid neutralization point is (0.000714 + 0.001286) = 0.0020 mole.
Total volume of the solution = (20 + 12.9) mL = 32.9 mL.
Molar concentration of HA at the acid equivalence point = (moles of HA/total volume of the solution) = (0.0020 mole)/(32.9 mL) = 6.079*10-5 mM.
HA establishes equilibrium as
HA ------> H+ + A-
Given Ka = 1.76*10-5, we have
Therefore, Ka = [H+][A-]/[HA] = (x).(x)/(6.079*10-5 – x)
Assume x is small (since Ka is small) and hence
1.76*10-5 = x2/(6.079*10-5)
===> x2 = 1.0699*10-9
===> x = 3.2709*10-5
Thus, pH = -log [H+] = -log (3.2709*10-5) =4.48 (ans).
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