You have prepared 250mL of a 0.200M acetate buffer solution with a pH of 4.53. b....

Question

Question

You have prepared 250mL of a 0.200M acetate buffer solution with a pH of 4.53.

Answer 1

Answer #1

total moles of buffer = 250 x 0.2 / 1000 = 0.05

acetic acid (x) + sodium acetate (y) = 0.05 ---------------------------> (1)

pH = pKa + log [y / x]

4.53 = 4.74 + log (y / x)

y/ x= 0.617

y = 0.617 x-----------------------> (2)

by solving 1 and 2

x = 0.0309

y = 0.0191

mass of sodium acetate = moles x molar mass

= 0.0191 x 136

= 2.6 g

volume of acetic acid = moles / molarity

= 0.0309 / 17.6

= 4.07 x 10^-3 L

= 4.07 mL

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Answer 2

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