《BUFFER SOLUTION》 1 What is a buffer solution

BUFFER SOLUTION DEPARTMENT OF MEDICAL CHEMISTRY Yongxiang Hu

BUFFER SOLUTION DEPARTMENT OF MEDICAL CHEMISTRY Yongxiang Hu

1. What is a buffer solution

1. What is a buffer solution

Example In I liter pure water add 0.01 mol of hydrochloric acid. Calculate this solution's pH change value. And in I liter contain 0.1 mol acetic acid and 0. 1 acetate ion mixture solution add 0.0lmol of hydrochloric acid. Calculate this solutions pH change value Notice: Square bracket will instead of equilibrium concentration in following time. Such as [H] instead hydrogen ion equilibrium concentration

Example: In 1 liter pure water add 0.01 mol of hydrochloric acid. Calculate this solution’s pH change value. And in 1 liter contain 0.1 mol acetic acid and 0.1 acetate ion mixture solution add 0.01mol of hydrochloric acid. Calculate this solution’s pH change value. Notice: Square bracket will instead of equilibrium concentration in following time. Such as [H+ ] instead hydrogen ion equilibrium concentration

Step 1. Calculate the pure water'S pH change value The pure waters hydrogen ion exponent (ph value (pure water)7.00 When add the hydrochloric acid TH=0.01mol/1 liter =0.01mol L-I pH=-og0.01=2.00 △pH=7.00-2.00=5.00( pH unit)

Step 1. Calculate the pure water’s pH change value The pure water’s hydrogen ion exponent (pH) value: pH(pure water) = 7.00 When add the hydrochloric acid: [H+ ] = 0.01mol/1 liter =0.01mol.L-1 pH = -log 0.01 = 2.00  pH = 7.00 – 2.00 = 5.00 (pH unit)

Step 2. Calculate contain acetic acid and its conjugate base solution's pH value In the solution acetic acid's concentration and acetate ions concentration HAc=0. 10 mol L- [Ac]=0.10 mol L-l Acetic acid equilibrium-constant Ka=1.70×105 In this solution has a acid-base equilibrium HAC Starting 0.10 0.10

Step 2. Calculate contain acetic acid and its conjugate base solution’s pH value In the solution acetic acid’s concentration and acetate ion’s concentration [HAc] = 0.10 mol.L-1 [Ac- ] = 0.10 mol.L-1 Acetic acid equilibrium-constant Ka = 1.7010-5 In this solution has a acid-base equilibrium HAc H+ + Ac￾Starting 0.10 0.10