What concentrations of acetic acid (pKa = 4.76) and acetate would be required to

Question

What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.10 M buffer solution at pH 4.6? Note that the concentration and/or pH value may differ from that in the first question. STRATEGY. Rearrange the Henderson-Hassel batch equation to solve for the ratio of base (acetate) to acid (acetic add), [A-] HA]. 2. Use the mole fraction of acetate to calculate the concentration of acetate. 3. Calculate the concentration of acetic acid Step 1 The ratio of base to acid to acid is [acetate]/ [acetic acid] = 0.69 Step 2: Use the mote fraction of acetate to calculate the concentration of acetate.

Explanation / Answer

From Henderson - Hasselbalch equation

pH = pKa + log [base]/[acid]

Given that pH = 4.6

pKa = 4.76

concentration of buffer solution = 0.10 M

For acetate-acetic acid buffer,

pH = pKa + log [acetate]/[acetic acid]

4.6 = 4.76 + log [acetate]/[acetic acid]

log [acetate]/[acetic acid] = 4.6-4.76

log [acetate]/[acetic acid] = -0.16

[acetate]/[acetic acid] = 10-0.16

[acetate]/[acetic acid] = 0.6918

[acetate] = [acetic acid] x 0.6918 -- Eq (1)

From given data, concentration of buffer solution = 0.10 M

i.e [acetate]+ [acetic acid] = 0.10 M ----Eq (2)

From Eq (1), [acetate] = [acetic acid] x 0.6918

Substitute this acetate value in Eq (2)

Then Eq (2) becomes, [acetic acid] x 0.6918 + [acetic acid] = 0.10 M

[acetic acid] { 0.6918+1 } = 0.10 M

[acetic acid] { 1.6918} = 0.10 M

[acetic acid] = 0.10 M / 1.6918

= 0.0591 M

[acetic acid] = 0.0591 M

[acetate]+ [acetic acid] = 0.10 M ----Eq (2)

then, [acetate]+ 0.0591 M = 0.10 M

  [acetate] = 0.10 M - 0.0591 M

= 0.0409 M

[acetate] =  0.0409 M

Therefore, [acetic acid] = 0.0591 M

[acetate] =  0.0409 M

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