Question

A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. You may want to reference (Pages 721 - 729) Section 17.2 while completing this problem. Part A What is the pH of this buffer? Express the pH to two decimal places. pHp H = nothing Request Answer Part B What is the pH of the buffer after the addition of 0.02 mol of NaOH? Express the pH to two decimal places. pHp H = nothing Request Answer Part C What is the pH of the buffer after the addition of 0.02 mol of HI? Express the pH to two decimal places.

Answer 1

Step-by-step explanation:

It is known that
`pK_(a)` of propionic acid = 4.87

And, initial concentration of propionic acid =
`(0.19)/(1.20)`

= 0.158 M

Concentration of sodium propionate = \frac{0.26}{1.20}[/tex]

= 0.216 M

Now, in the given situation only propionic acid and sodium propionate are present .

Hence, pH =
`pK_(a) + log(([salt])/([acid]))`

=
`4.87 + log (0.216)/(0.158)`

= 4.87 + log (1.36)

= 5.00

• Therefore, when 0.02 mol NaOH is added then,

Moles of propionic acid = 0.19 - 0.02

= 0.17 mol

Hence, concentration of propionic acid =
`(0.17)/(1.20 L)`

= 0.14 M

and, moles of sodium propionic acid = (0.26 + 0.02) mol

= 0.28 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

`(0.28 mol)/(1.20 L)`

= 0.23 M

Therefore, calculate the pH upon addition of 0.02 mol of NaOH as follows.

pH =
`pK_(a) + log(([salt])/([acid]))`

=
`4.87 + log (0.23)/(0.14)`

= 4.87 + log (1.64)

= 5.08

Hence, the pH of the buffer after the addition of 0.02 mol of NaOH is 5.08.

• Therefore, when 0.02 mol HI is added then,

Moles of propionic acid = 0.19 + 0.02

= 0.21 mol

Hence, concentration of propionic acid =
`(0.21)/(1.20 L)`

= 0.175 M

and, moles of sodium propionic acid = (0.26 - 0.02) mol

= 0.24 mol

Hence, concentration of sodium propionic acid will be calculated as follows.

`(0.24 mol)/(1.20 L)`

= 0.2 M

Therefore, calculate pH upon addition of 0.02 mol of HI as follows.

pH =
`pK_(a) + log(([salt])/([acid]))`

=
`4.87 + log (0.2)/(0.175)`

= 4.87 + log (0.114)

= 4.98

Hence, the pH of the buffer after the addition of 0.02 mol of HI is 4.98.