Question

ignment Score: 490/1000 Resources Lt Give Up? Feedback Try Again Question 6 of 10 > Attempt 2 You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (pK= 4.20) and 0.200 M sodium benzoate. buffer? How many milliliters of each solution should be mixed to prepare this buffer? benzoic acid: 65.98 sodium benzoate: 34.02 (hp) BANGL OLUFSEN

Answer 1

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Let $V_{1}$ be the volume of sodium benzoate solution .

Given that, Total volume is 100 mL.

Therefore, Volume of benzoic acid,

V2 = 100 – 11

Given that,

Molarity of sodium benzoate solution,

Mi = 0.2 M

Molarity of benzoic acid,

M = 0.1 M

pH = 4

pKa of benzoic acid,

pka = 4.2

According to Henderson-Hasselbalch equation, we have,

pH = pka + log [salt] Jacid]

, 4 = 4.2+108 [ [ , 4 _ 4 21 ( [ᏁᎷᏙi] Ꮩ2

[0.2 x V1] ::-0.2 = logo 0.1 x (100 – V1)]

[0.2 x V) 10.1 x (100 - 12) = 0.630957

Solving for $V_{1}$ , we get

V1= 23.98 mL

::V= 100 ml - 23.98 ml

:: V2= 76.02 mL

That is volume of benzoic acid is
76.02 mL
and volume of sodium benzoate solution is
23.98 mL.

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