Question

A buffer contains 0.19 mol of propionic acid (C2H5COOH) and 0.22 mol of sodium propionate (C2H5COONa) in 1.20 L. Ka = 1.3 x 10^-5

Part A

What is the pH of this buffer?

Part B

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

Part C

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

Answer 1

A)

Ka = 1.3*10^-5

pKa = - log (Ka)

= - log(1.3*10^-5)

= 4.886

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.886+ log {0.22/0.19}

= 4.95

Answer: 4.95

B)

mol of NaOH added = 0.02 mol

C2H5COOH will react with OH- to form C2H5COO-

Before Reaction:

mol of C2H5COO- = 0.22 mol

mol of C2H5COOH = 0.19 mol

after reaction,

mol of C2H5COO- = mol present initially + mol added

mol of C2H5COO- = (0.22 + 0.02) mol

mol of C2H5COO- = 0.24 mol

mol of C2H5COOH = mol present initially - mol added

mol of C2H5COOH = (0.19 - 0.02) mol

mol of C2H5COOH = 0.17 mol

Ka = 1.3*10^-5

pKa = - log (Ka)

= - log(1.3*10^-5)

= 4.886

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.886+ log {0.24/0.17}

= 5.036

Answer: 5.04

C)

mol of HI added = 0.02 mol

C2H5COO- will react with H+ to form C2H5COOH

Before Reaction:

mol of C2H5COO- = 0.22 mol

mol of C2H5COOH = 0.19 mol

after reaction,

mol of C2H5COO- = mol present initially - mol added

mol of C2H5COO- = (0.22 - 0.02) mol

mol of C2H5COO- = 0.2 mol

mol of C2H5COOH = mol present initially + mol added

mol of C2H5COOH = (0.19 + 0.02) mol

mol of C2H5COOH = 0.21 mol

Ka = 1.3*10^-5

pKa = - log (Ka)

= - log(1.3*10^-5)

= 4.886

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.886+ log {0.2/0.21}

= 4.865

Answer: 4.86