In the Hydrogen Ion Equilibrium experiment, we will be performing a titration of
ID: 899568 • Letter: I
Question
In the Hydrogen Ion Equilibrium experiment, we will be performing a titration of a solution containing 250 mL of imidazolium, ethanolammonium, triethylammonium cations, and EBT (EBT. at 2-3 times 10^-5 M, is just going along for the ride) with 0.1M KOH. The purpose of the titration is to determine the concentration (spectroscopically) of the relevant EBT species present at a given pH. We will start with a solution containing primarily H_2D at a pH of 4. As KOH is added, it reacts with the major acid Species in the buffer, thereby changing the pH. As the pH changes, so do the relative concentrations of H_2D to HD^2 (pH near pK a) and HD^2- D^ 3-(pH near Pka2 We want to have enough points in order to get a smooth looking titration curve, so you'll need to get at least 7-8 points in the range of 20-80% conversion of H_2D- to HD^2,2-3 points near pH 8.8-9.8 where HD" is present at about 99% and 6-7 points at 20-60% conversion of HD^2 to D^3. What values do Diehl and Lindstrom give for pkal and pK a2 of EBT? What is the pH at which 20% of H_2D- is converted to HD2-?80%? Use the Henderson-Hasselbalch equation with pKa1to find the Ph. What pH values correspondent to 20 and 60% conversion of HD^2- to D^3-? Use the Henderson - Hasselbalch equation with pKa2 to find the pH.Explanation / Answer
a) using the method, we get
pKa1= 6.3
pKa2=12.4
b)when 20% is converted,
pH=pKa2 + log(salt/acid)
=6.3 + log(20/80)
=5.697
when 80% is converted,
pH=pKa2 + log(salt/acid)
=6.3 + log(80/20)
=6.9
c) when 20 % is converted,
pH=pKa2 + log(salt/acid)
=12.4 + log(20/80)
=11.79
when 60% is converted,
pH=pKa2 + log(salt/acid)
=12.4 + log(60/40)
=12.57