Determine the volume increments of 0.10 M KOH needed to reach and then span the

Question

Determine the volume increments of 0.10 M KOH needed to reach and then span the 20-80% H_2D HD^2 region (250 mL 0.005 M imidazolium cation, pKa, 7.0, 250 mL 0.0025 M ethanolammonium, pKa 9.5) and to reach the HD^2 region. Use the pHs calculated in Part 1 (i.e, 20% conversion of H_2D^- to HD^2-) and the Henderson-Hasselbalch equation to determine the ratio of conjugate base to acid for each of the species in the buffer: Imidazolium cation = Him,Im^- = conjugate base, pKa = 7.0, to find the ratio of Im^- /HIm. Ethanolammonium cation HEt, Et = conjugate base, pKa = 9.5, to find the ratio of Et^- /HEt. Since you know how much of each cation you have in the buffer initially, calculate the volume of KOH required to generate the appropriate ratio. Determine the volume increments of 0.10 M KOH needed to reach and then span the 20-60% HD^2- _D^3- region (rest of the ethanolammonium and 250 ml. 0.005 M triethylammonium, pKa 10.8, and. especially above pH 11, where most of the pKa2 action is, the KOH needed just to raise the pH of 250 mL of pure water). Employ the same set of calculations as in part 2.a. with HEt/Et and H(Et)3/(Et)3.

Explanation / Answer

(a) For Imidazolium cation: Moles of imidazolium cation = MxV = 0.005Mx0.250L = 0.00125 mol

Concentration of KOH = 0.10M

Let the volume of KOH added = V mL

Here we need 20 % conversion of imidazolium cation to imidazole.

Hence 20% of  imidazolium cation = 0.00125 mol x (20/100) = 0.00025 mol

The chemical equation for the reaction of Imidazolium cation(HIm) with KOH is

HIm + OH- ---- > Im- + H2O

1 mol 1 mol 1 mol

In the above reaction, 1 mol of HIm reacts with 1 mol of KOH.

Hence 0.00025(20%) mol of HIm that will react with the moles of KOH = 0.00025 mol KOH.

Also 0.00025 mol KOH = 0.10xV

=> V = 0.00025 / 0.10 L = 0.0025 L = 2.5 mL

Hence volume of KOH required = 2.5 mL

For ethanolammonium cation: Moles of ethanolammonium cation = MxV = 0.0025Mx0.250L = 0.000625 mol

Concentration of KOH = 0.10M

Let the volume of KOH added = V mL

Here we need 20 % conversion of ethanolammonium cation to ethanolamonia.

Hence 20% of ethanolammonium cation = 0.000625 mol x (20/100) = 0.000125 mol

The chemical equation for the reaction of ethanolammonium cation(HEt) with KOH is

HEt + OH- ---- > Et- + H2O

1 mol 1 mol 1 mol

In the above reaction, 1 mol of HEt reacts with 1 mol of KOH.

Hence 0.000125(20%) mol of HEt that will react with the moles of KOH = 0.000125 mol KOH.

Also 0.000125 mol KOH = 0.10xV

=> V = 0.000125 / 0.10 L = 0.00125 L = 1.25 mL

Hence volume of KOH required = 1.25 mL

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