Question

4 35mL of 2.5M HCI is added to 1.0L of the formic acid buffer prepared previously (0.20M HCHO and 0.15M NaCHO2). What is the pH of the resulting solution? Ka- 1.8x10

Answer 1

mol of HCl added = 2.5M *0.035 L = 0.0875 mol

CHO2- will react with H+ to form HCHO2

Before Reaction:

mol of CHO2- = 0.15 M *1.0 L

mol of CHO2- = 0.15 mol

mol of HCHO2 = 0.2 M *1.0 L

mol of HCHO2 = 0.2 mol

after reaction,

mol of CHO2- = mol present initially - mol added

mol of CHO2- = (0.15 - 0.0875) mol

mol of CHO2- = 0.0625 mol

mol of HCHO2 = mol present initially + mol added

mol of HCHO2 = (0.2 + 0.0875) mol

mol of HCHO2 = 0.2875 mol

Ka = 1.8*10^-4

pKa = - log (Ka)

= - log(1.8*10^-4)

= 3.745

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

use:

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

= 3.745+ log {6.25*10^-2/0.2875}

= 3.082

Answer: 3.08