2) Consider a buffer solution made from a partially neutralizing 1.00 M hydrochl
ID: 1020996 • Letter: 2
Question
2) Consider a buffer solution made from a partially neutralizing 1.00 M hydrochlorious acid (HClO, Ka=3.5*10^-8) with 0.500M sodium hydroxide.
A Calculate the amount of base needed to make a buffer from 50.0mL of HClO with a pH equal to the pKa of the acid.
B What is the pH of a buffer solution made in (a) after diluting to 500mL.
C) Which solution made in (a) or (b) will have a lower capacity if equal volumes of each are compared? Why ?
D) Calculate the pH of a buffer made using 30.0mL of the acid and 10.0mL of the base.
E)Calculate the pH of a buffer using 20.0ml of acid and 30.0mL of base.
F) Which solution made in (d) or (e) can buffer more acid ? Which can buffer more base? Why?
Explanation / Answer
2) Consider a buffer solution made from a partially neutralizing 1.00 M hydrochlorious acid (HClO, Ka=3.5*10^-8) with 0.500M sodium hydroxide.
A Calculate the amount of base needed to make a buffer from 50.0mL of HClO with a pH equal to the pKa of the acid.
pKa = - log Ka = -log 3.5 x 10-8
pKa = 7.45
So base = acid when pH = pKa
For base to be equal to acid we should add half base as acid as acid will be consumed to form the conjugate base
So base needed is 50 mL of 0.5 M NaOH
B What is the pH of a buffer solution made in (a) after diluting to 500mL.
pH does not chnage by dilution so pH = 7.45
C) Which solution made in (a) or (b) will have a lower capacity if equal volumes of each are compared? Why ?
Solution (a) will have more capacity as capacity is dependent on concentration of base and acid since (a) has higher concentration it has more capacity.
D) Calculate the pH of a buffer made using 30.0mL of the acid and 10.0mL of the base.
Total volume is 30 + 10 = 40 mL
molarity of base =
10 x 0.5 = 40 x N2
N2 = 0.125 M
30mL of acid is
30 x 1 = 40 x N2
N2 = 0.75 M - 0.125 M = 0.625 M acid as 0.125 M is consumed in forming the conjugate base
pH = pKa + log base/acid
pH = 7.45 + log 0.125/0.625
pH = 6.75
E)Calculate the pH of a buffer using 20.0ml of acid and 30.0mL of base.
Total volume is 20 + 30 = 50 mL
molarity of base =
30 x 0.5 = 50 x N2
N2 = 0.3 M
20mL of acid is
20 x 1 = 50 x N2
N2 = 0.4 M - 0.3 M = 0.1 M acid as 0.3 M is consumed in forming the conjugate base
pH = pKa + log base/acid
pH = 7.45 + log 0.3/0.1
pH = 7.93
F) Which solution made in (d) or (e) can buffer more acid ? Which can buffer more base? Why?
(e) can buffer more acid as the ratio of base to acid is high and (e) can buffer more base as ratio of base/acid is low as by adding acid to (e) you can shift the ratio of base to acid to being more positive and the reverse for (d)
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