You need to produce a buffer solution that has a pH of 5.40. You already have a solution that contains 10. mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74. Express your answer numerically in millimoles. View Available Hint(s)

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znk znk · Answer 1 · 2018-12-26T17:19:00+01:00

Answer:

46. mmol

Step-by-step explanation:

The equation for the equilibrium is:

HA + H₂O ⇌ A⁻ + H₃O⁺

The solution is a buffer, so we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A⁻]/[HA])

5.40 = 4.74 + log([A⁻]/(10.) Subtract 4.74 from each side

0.66 = log([A⁻]/(10.) Take the antilog of each dide

[A⁻]/10. = 10 ^0.66

[A⁻]/10. = 4.57 Multiply each side by 10.

[A⁻] = 46. mmol

You will add 46. mmol of sodium acetate.

batolisis batolisis · Answer 2 · 2021-10-14T15:32:50+02:00

The amount of acetate needed to add to the solution is ; 46 mmol

Given data:

pH of buffer solution = 5.4

volume of Solution ( H A ) = 10 mmol

pKa of acetic acid = 4.74

Given that the solution is a buffer solution we will apply Henderson-Hasselbalch equation:

pH = [tex]pKa + log ( [A^-] / [HA] )[/tex] -------- ( 1 )

Insert values into equation 1

[tex]5.40 = 4.74 + log([A^-]/(10 )[/tex]

0.66 = [tex]log([A^-]/(10.)[/tex] -------- ( 2 ) ( after subtracting pKa value from both side )

∴ [tex][A^-]/10. = 10^{0.66}[/tex] ----- ( 3 ) ( antilog )

Multiply both sides of equation 3 by 10

[tex][A^-] = 46 mmol[/tex].

Hence we can conclude that the amount of acetate needed to add to the solution is 46 mmol

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