Question

a. Acetic acid (CHCOH, K = 1.8 10) and sodium acetate (NaCHO).

b.Ammonium chloride (NHCl, K for NH = 5.6 10) and ammonia (NH).

Given: pH = 4.5 and 0.45M.
1.Decide which of the two available buffer systems you should use.

2.Determine the ratio of [A]/[HA] needed in the buffer.

3.Which component will be more concentrated, A or HA? Choose a value for the concentration of that component anywhere in the range 0.2 – 0.6 M. Calculate what the concentration of the other component should be.

4.Calculate the volume (in mL) of 17.5 M CH3CO2H or 14.8 M NH3 needed to make 200 mL of your buffer. Calculate the mass (in g) of solid NaCH3CO2·3H2O or NH4Cl needed to make 200 mL of your buffer.
Please help. Thanks,

Answer 1

Acetic acid (CH₃CO₂H): K_a = 1.8*10^-5

Ammonium chloride (NH₄Cl): K_a (NH₄⁺) = 5.6*10^-10

We need to prepare a buffer at pH 4.5 and total buffer concentration of 0.45 M. Start by finding out the pK_as of the two buffer systems mentioned.

1) pK_a (CH₃CO₂H) = -log K_a = -log (1.8*10^-5) = 4.74

pK_a (NH₄⁺) = -log K_a = -log (5.6*10^-10) = 9.25

For a buffer system to effectively perform its buffering action at a particular pH, the pK_a of the buffer must be within 1 unit of the given pH. Since the desired pH is 4.5 and CH₃CO₂H has a pK_a value of 4.74, therefore, choose the CH₃CO₂H/NaCH₃CO₂ as the appropriate buffer (acetic acid/acetate buffer).

2) CH₃CO₂H dissociates as

CH₃CO₂H (aq) <=======> CH₃CO₂^- + H⁺

Let us denote acetic acid as HA and acetate base as A^-. Use the Henderson-Hasslebach equation to calculate the molar ratio of the conjugate base and the acid.

pH = pK_a + log [A^-]/[HA]

Plug in pH = 4.5 and pK_a = 4.74 and obtain

4.5 = 4.74 + log [A^-]/[HA]

===> -0.24 = log [A^-]/[HA]

===> [A^-]/[HA] = antilog (-0.24) = 0.575

Therefore, [A^-]/[HA] = 0.575/1; thus the molar ratio of the conjugate base (acetate) and the weak acid (acetic acid) is 0.575:1 (ans).

3) Since the ratio of [A^-] and [HA] is less than 1, hence HA will be more concentrated and A^- will be less concentrated (ans).

Next, find out the concentration of HA and A^- in the buffer using the relation

[A^-] + [HA] = 0.45 (given the total buffer strength is 0.45 M)

===> 0.575*[HA] + [HA] = 0.45 (since [A^-]/[HA] = 0.575)

===> 1.575*[HA] = 0.45

===> [HA] = 0.45/1.575 = 0.2857 ≈ 0.286

Therefore, [A^-] = 0.45 – [HA] = 0.45 – 0.286 = 0.164

Thus the concentrations of HA and A^- in the buffer are 0.286 M and 0.164 M respectively (ans).

4) We wish to prepare 200 mL of buffer solution having [HA] = 0.286 M and [A^-] = 0.164 M.

Available stock solution is 17.5 M HA.

Use the dilution equation

M₁*V₁ = M₂*V₂ where M₁ = 17.5 M; M₂ = 0.286 M and V₂ = 200 mL.

Plug in values and write

(17.5 M)*V₁ = (0.286 M)*(200 mL)

===> V₁ = (0.286*200)/(17.5) mL = 3.268 mL ≈ 3.27 mL.

Pipette out 3.27 mL of 17.5 M acetic acid into a 200 mL volumetric flask and make up the volume with water (ans).

Find the moles of acetate base present in the buffer = (volume of buffer in L)*(concentration of acetate in the buffer) = (200 mL)*(1 L/1000 mL)*(0.164 mol/L) = 0.0328 mole.

Molar mass of sodium acetate = 82.0343 g/mol.

Therefore, mass of sodium acetate to be weighed out = (moles of acetate)*(molar mass of sodium acetate) = (0.0328 mole)*(82.0343 g/mol) = 2.6907 g ≈ 2.691 g

Weigh out 2.691 g of solid sodium acetate and add to the volumetric flask, shake and dissolve and finally make up the volume to prepare the buffer (ans).