Question

1. When 0.0075 moles of HCl are added to 33 mL of a buffer that is 0.432 M HX and 0.876 M NaX the pH is measured to be 4.81. What is the K_aof the acid, HX?

Answer 1

initial moles of HX = 0.432 x 33 /1000 = 0.0142

initial moles of NaX = 0.876 x 33 / 1000 = 0.0289

moles of HCl added = 0.0075

after HCl added

moles of HX = 0.0142 - 0.0075 =0.0217

moles of NaX = 0.0289 - 0.0075 = 0.0214

[HX] = 0.0217 / 0.033 = 0.66 M

[NaX] = 0.0214 / 0.033 = 0.65 M

pH = 4.81

pH = pKa + log [NaX] / [HX]

4.81 = pKa + log [0.65] / [0.66]

4.81 = pKa - 0.0066

pKa = 4.817

now

Ka = 10^-pKa = 10^-4.817

Ka = 1.52 x 10^-5