You need to prepare 100.0 mL of a pH=4.00 buffer solution using 0.100 M benzoic acid...

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You need to prepare 100.0 mL of a pH=4.00 buffer solution using0.100 M benzoic acid (pKa = 4.20) and 0.220 M sodium benzoate. Howmuch of each solution should be mixed to prepare this buffer?
a) mL of benzoic acid
b) mL of sodium benzoate

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4.0 Ratings (815 Votes)

Let volume of sodium benzoate =x

Â Â Â Â Â volume of benzoic acidÂ Â Â Â Â = 100-x

Â Â Â Â Â no of moles of sodium benzoate = molarity * volume

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.22*x

Â Â Â Â no of moles of bezoic acidÂ Â Â Â Â Â Â Â Â Â Â = molarity * volume

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.1*(100-x)

Â Â Â Â Â Â PH = PKa +log[sodium benzoate]/[benzoic acid]

Â Â Â Â Â Â 4 Â Â Â Â = 4.2 + log0.22*x/0.1*(100-x)

Â Â Â Â Â Â 4-4.2Â Â = log0.22*x/0.1*(100-x)

Â Â Â Â Â Â -0.2Â Â Â Â = log0.22*x/0.1*(100-x)

log0.22*x/0.1*(100-x) = -0.2

0.22*x/0.1*(100-x)Â Â Â Â Â Â = 10^-0.2 = 0.63

0.22*x/0.1*(100-x)Â Â = 0.63

0.22*x Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.63*0.1(100-x)

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.063(100-x)

0.22x Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =6.3-0.063x

0.22x-6.3+0.063x =0

0.263x-6.3 =0

Â Â 0.263x = 6.3

Â Â Â Â Â Â Â Â xÂ Â = 6.3/0.263Â Â = 23.95

volume of benzoic acidÂ Â = 23.95ml

volume of benzoateÂ Â Â Â Â Â = 100-x = 100-23.95 = 76.05ml

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