Question

A total of 0.075 moles (=75 mmol) of HCl is added to 200.0mL of a buffer that contains a 0.25M HC7H5O2 and 0.50M NaC7H5O2. What is the final pH of this solution?? Ka=6.5x10^-5 pka=4.187

I know the final answer is pH= 3.49

But I can never get that answer exactly.

Please show steps!!

Answer 1

mol of HCl added = 75.0 mmol

C6H5COONa will react with H+ to form C6H5COOH

Before Reaction:

mol of C6H5COONa = 0.5 M *200.0 mL

mol of C6H5COONa = 100 mmol

mol of C6H5COOH = 0.25 M *200.0 mL

mol of C6H5COOH = 50 mmol

after reaction,

mol of C6H5COONa = mol present initially - mol added

mmol of C6H5COONa = (100 - 75.0) mmol

mol of C6H5COONa = 25 mmol

mol of C6H5COOH = mol present initially + mol added

mol of C6H5COOH = (50 + 75.0) mmol

mol of C6H5COOH = 125 mmol

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

use:

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

= 4.187+ log {25/125}

= 3.49

pH is 3.49