Question

A 1.0-L buffer solution contains 0.100 molHC2H3O2 and 0.100 molNaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5. You may want to reference (Pages 648 - 658) Section 16.2 while completing this problem. Part A Calculate the pH of the solution upon addition of 48.1 mL of 1.00 MHCl to the original buffer.

Answer 1

mol of HCl added = 1.0M *0.0481 L = 0.0481 mol

C2H3O2- will react with H+ to form HC2H3O2

Before Reaction:

mol of C2H3O2- = 0.1 mol

mol of HC2H3O2 = 0.1 mol

after reaction,

mol of C2H3O2- = mol present initially - mol added

mol of C2H3O2- = (0.1 - 0.0481) mol

mol of C2H3O2- = 0.0519 mol

mol of HC2H3O2 = mol present initially + mol added

mol of HC2H3O2 = (0.1 + 0.0481) mol

mol of HC2H3O2 = 0.1481 mol

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

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

use:

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

= 4.745+ log {5.19*10^-2/0.1481}

= 4.289

Answer: 4.29