Question

A buffer contains 0.11 mol of propionic acid (C2H5COOH) and 0.25 mol of sodium propionate (C2H5COONa)...

A buffer contains 0.11 mol of propionic acid (C2H5COOH) and 0.25 mol of sodium propionate (C2H5COONa) in 1.20 L.

A:

What is the pH of this buffer?

Express the pH to two decimal places.

B:

What is the pH of the buffer after the addition of 0.02 mol of NaOH?

Express the pH to two decimal places.

C:

What is the pH of the buffer after the addition of 0.02 mol of HI?

Express the pH to two decimal places.

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Answer #1

Buffer solution is made by the addition of acid and base solution. The pH of this solution is remain unchanged by adding acid or base.   In general the pH of this solution is constant. . A buffer must consist of a weak conjugate acid-base pair, thus it is a weak acid and its conjugate base, or a weak base and its conjugate acid.

Ka = 1.3x10^-5

Therefore pKa= log Ka

= log 1.3x10^-5

= 4.87

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

=4.87 + log 0.25/ 0.11

= 4.87 + log 2.27

= 4.87+ (0.367)

= 5.24

0.02 mol of NaOH

HC3H5O2(aq) + NaOH(aq) ----> NaC3H5O2(aq) + H2O(l)

moles acid = 0.11 - 0.02 = 0.09
[acid]= 0.09/ 1.20=0.075 M
moles salt = 0.25 + 0.02 = 0.27
[salt]= 0.27/ 1.20=0.225


pH = 4.87 + log 0.225/ 0.075 = 5.35

0.02 mol of HI
moles acid = 0.11 +0.02 = 0.13
[acid]= 0.13/ 1.20=0.108 M
moles salt = 0.25 - 0.02 = 0.23
[salt]= 0.23/ 1.20=0.192

pH = 4.87 + log 0.192/ 0.108= 5.12

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