A buffer is prepared such that [CH 3 COOH] 0 = [CH 3 COO – ] 0 = 0.100 M. What v
ID: 945565 • Letter: A
Question
A buffer is prepared such that [CH3COOH]0 = [CH3COO–]0 = 0.100 M. What volume of 0.125 M NaOH (aq) can be added to 200.0 mL of this buffer solution before its buffering capacity is lost. Assume that the buffer capacity is lost when the ratio [base]0/[acid]0 is less than 0.1 or greater than 10.
Select one:
a. 123 mL
b. 145 mL
c. 113 mL
d. 131 mL
Question 2
Given a solution of 0.100 M benzoic acid, C6H5COOH (aq), and a solution of 0.150 M NaC6H5COO (aq), how would you prepare 100.0 mL of a buffer solution with a pH of 4.35? pKa = 4.20 for C6H5COOH (aq).
Select one:
a. Mix 54.0 mL of the C6H5COOH (aq) with 46.0 mL of the NaC6H5COO (aq).
b. Mix 50.0 mL of the C6H5COOH (aq) with 50.0 mL of the NaC6H5COO (aq).
c. Mix 51.5 mL of the C6H5COOH (aq) with 48.5 mL of the NaC6H5COO (aq).
d. Mix 48.0 mL of the C6H5COOH (aq) with 52.0 mL of the NaC6H5COO (aq).
Explanation / Answer
pH = pka + log(salt/acid)
if the [CH3COOH]0 = [CH3COO–]0 = 0.100 M
pH = pka = 4.74
by the addition of NaOH pH increases,maximum buffer capacity up to 5.74
pH = 5.74
No of mol of buffer = 0.2*0.1 = 0.02 mol
pH = pka + log(salt+x/acid-x)
5.74 = 4.74 + log((0.02+x)/(0.02-x))
x = 0.01636 mol
Volume of NaOH = n/M = 0.01636/0.125 = 0.13088 L
= 130.88 ml
answer: d.
2) pH = pka + log(salt/acid)
4.35 = 4.2+log(x)
x= 1.41
so that
x = salt/acid = 1.41
form the options
C option is having ratio = (48.5*0.15)/(0.1*51.5) = 1.41
answer: C
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.