Question

In: Chemistry

You are asked to prepare 500. mL of a 0.300 M acetate buffer at pH 4.90...

You are asked to prepare 500. mL of a 0.300 M acetate buffer at pH 4.90 using only pure acetic acid (MW=60.05 g/mol, pKa=4.76), 3.00 M NaOH, and water. Answer the following questions regarding the preparation of the buffer.

What volume of 3.00 M NaOH must you add to the acetic acid to achieve a buffer with a pH of 4.90 at a final volume of 500 mL? (Ignore activity coefficients.)

Expert Solution

PH = PKa + log[NaAc]/[HAc]

4.9 = 4.76 + log[NaAc]/[HAc]

log[NaAc]/[HAc] = 4.9-4.76

log[NaAc]/[HAc] = 0.14

[NaAc]/[HAc] = 10^0.14

[NaAc]/[HAc] = 1.38

[NaAc] = 1.38[HAc]

no of moles of acetic acid (HAc) = molarity * volume in L

= 0.3*0.5 = 0.15moles

total no of moles in buffer = 0.15moles

[NaAc]+[HAc] = 0.15

1.38[HAc]+[HAc] = 0.15

2.38[HAc] = 0.15

[HAc] = 0.15/2.38

[HAc] = 0.063moles

[NaAC] = 0.15-0.063 = 0.087moles

NaOH react with HAc to gives NaAC 0.087 moles of NaOH is needed

no of moles of NaOH = 0.087 moles

molarity of NaOH = no of moles/volume in L

3 = 0.087/volume in L

volume in L = 0.087/3 = 0.029L = 29ml

queen_honey_blossom answered 1 year ago

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