You are to assume that the buffer is made up by mixing volumes of 0.100 M acetic

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.

The volume of 0.100 M acetic acid required to prepare 60.0 mL of a buffer of pH 4.8. = 28.1 mL

The volume of 0.100 M sodium acetate required to prepare 60.0 mL of a buffer of pH 4.8. = 31.9 mL

The concentration [HA] in this buffer. = 0.0468 M

The concentration [A-] in this buffer. = 0.0532 M

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. = 9.4 mL

Calculate the pH at the Basic equivalence point.

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. = 10.6 mL

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) = (0.0468 mol/L)*(20 mL)*(1 L/1000 mL) = 0.000936 mole.

Moles of A- in 20 mL of the buffer solution = (molar concentration of A-)*(volume of buffer solution in L) = (0.0532 mol/L)*(20 mL)*(1 L/1000 mL) = 0.001064 mole.

The basic neutralization reaction is

HA + NaOH -----> NaA + H2O

Moles of NaOH added = moles of HA completely neutralized = 0.000936 mole.

Molar concentration of NaOH = 0.1000 M.

Therefore, volume of NaOH required = (moles of HA neutralized)/(molar concentration of NaOH) = (0.000936 mole)/(0.1000 mol/L) = 0.00936 L = (0.00936 L)*(1000 mL/1 L) = 9.36 mL 9.4 mL (ans).

At the basic neutralization point, we have 0.000936 mole of HA completely neutralized to form 0.000936 mole of A-. Therefore, the total number of moles of A- at the basic neutralization point is (0.000936 + 0.001064) = 0.0020 mole.

Total volume of the solution = (20 + 9.4) mL = 29.4 mL.

Molar concentration of A- at the basic equivalence point = (moles of A-/total volume of the solution) = (0.0020 mole)/(29.4 mL) = 6.803*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/(6.803*10-5)

===> x2 = 3.864*10-14

===> x = 1.966*10-7

Therefore, [OH-] = 1.966*10-7 and pOH = -log [OH-] = -log (1.966*10-7) = 6.706

Thus, pH = 14 – pOH = 14 – 6.706 = 7.294 7.3 (ans).

The acid neutralization reaction is

A- + HCl -----> HA + Cl-

Moles of HCl added = moles of A- completely neutralized = 0.001064 mole.

Molar concentration of HCl = 0.1000 M.

Therefore, volume of HCl required = (moles of A- neutralized)/(molar concentration of HCl) = (0.001064 mole)/(0.1000 mol/L) = 0.001064 L = (0.001064 L)*(1000 mL/1 L) = 10.64 mL 10.6 mL (ans).

At the acid neutralization point, we have 0.001064 mole of A- completely neutralized to form 0.001064 mole of HA. Therefore, the total number of moles of HA at the acid neutralization point is (0.000936 + 0.001064) = 0.0020 mole.

Total volume of the solution = (20 + 10.6) mL = 30.6 mL.

Molar concentration of HA at the acid equivalence point = (moles of HA/total volume of the solution) = (0.0020 mole)/(30.6 mL) = 6.536*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)/(6.536*10-5 – x)

Assume x is small (since Ka is small) and hence

1.76*10-5 = x2/(6.536*10-5)

===> x2 = 1.150*10-9

===> x = 3.392*10-5

Thus, pH = -log [H+] = -log (3.392*10-5) =4.469 4.47 (ans).

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