You need to prepare 100.0 mL of a pH 4.00 buffer solution using 0.100 M benzoic acid (p K a = 4.20 ) and 0.220 M sodium benzoate. How many milliliters of each solution should be mixed to prepare this buffer?

Volume benzoic acid= 77.74 mL  

Volume sodium benzoate: = 22.26 mL

Explanation:

Step 1: Data given

Volume = 100.0 mL

pH = 4.00

Molarity benzoic acid = 0.100 M

pKa benzoic acid = 4.20

Molarity sodium benzoate = 0.220 M

Step 2

pH = pKa + log([base]/[acid])

4 = 4.2 + log([base]/[acid])

-0.2 = log([base]/[acid])

10^-0.2 = [base]/[acid]  

0.63 =  [base]/[acid]  

Step 3

start with 100 mL 0.1 M benzoic acid

0.100 L* 0.1M = 0.01 moles.  

Since base/acid = 0.63, add 0.63*0.01 moles base = 0.0063 moles base.

And 0.0063 moles of a 0.22 M salt solution are x mL:

0.22 moles = 1000 mL

1 mole = 1000/0.22 mL

0.0063 moles = 1000*0.0063/0.22 mL = 28.63 mL .

So a mixture of 100 mL 0.1 M bencoic acid + 28.63 mL of 0.22 benzoate has pH =4.00

Step 4: For a total volume of 100 mL we'll have:

Volume benzoic acid: 100*100/(128.63) = 77.74 mL  

Volume sodium benzoate: 28.63*100/(128.63) = 22.26 mL

b is the correct answer

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i think that the answer is c