Question

A 1.0 L buffer solution contains 0.192 MHC2H3O2 and 0.192 M NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5. Because the initial amounts of acid and conjugate base are equal, the pH of the buffer is equal to pKa=−log(1.8×10−5)=4.74.

Calculate the new pH after adding 0.019 mol of solid NaOH to the buffer.

Answer 1

mol of NaOH added = 0.019 mol

HC2H3O2 will react with OH- to form C2H3O2-

Before Reaction:

mol of C2H3O2- = 0.192 M *1.0 L

mol of C2H3O2- = 0.192 mol

mol of HC2H3O2 = 0.192 M *1.0 L

mol of HC2H3O2 = 0.192 mol

after reaction,

mol of C2H3O2- = mol present initially + mol added

mol of C2H3O2- = (0.192 + 0.019) mol

mol of C2H3O2- = 0.211 mol

mol of HC2H3O2 = mol present initially - mol added

mol of HC2H3O2 = (0.192 - 0.019) mol

mol of HC2H3O2 = 0.173 mol

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.744727

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

use:

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

= 4.744727+ log {0.211/0.173}

= 4.83

pH is 4.83