CHEMISTRY 31

Summer, 2016 - Dixon

Additional Problem 3.2 - Solutions

Calculate the volume of 1.00 M HCl and the mass of 2-aminoethanol needed to make 250.0 mL of a pH = 9.20 buffer with a total 2-aminoethanol concentration of 0.0100 M. Total 2-aminoethanol is the sum of 2-aminoethanol and its conjugate acid. Use Appendix G to find relavant K_a values.

2-aminoethanol is aweak base, so addition of HCl, a strong acid, will convert some of 2-aminoethanol to its conjugate weak acid:

B(aq) + H⁺ ↔ BH⁺

To get a total 2-aminoethanol concentration of 0.0100 M, we can add enough moles of 2-aminoethanol. n(2-aminoethanol) = (0.0100 mol/L)(250 mL)(1L /1000 mL) = 0.00250 moles.

Needed mass = (0.00250 mol)(61.09 g/mol) = 0.153 g.

Next, we can create a mole table to determine the moles of HCl needed to convert the right amount of 2-aminoethanol (=B) to its conjugate acid.

B(aq) + H⁺ ↔ BH⁺

Initial 0.00250 x 0

Change -x -x +x

Full right 0.00250 – x 0 x

Now, we can use the Henderson-Hasselbalch equation:

pH = pK + log[B]/[BH⁺]

or 9.20 = 9.498 + log[(0.00250 – x)/x] or -0.298 = log[(0.00250 – x)/x]

0.5035 = (0.00250 – x)/x or 0.5035x = 0.0025 – x or 1.5035x = 0.00250

X = 0.00250/1.5035 = 0.00166 mol = [HCl]V(HCl)

V(HCl) = 0.00166 mol/1.00 mol/L = 0.00166 L = 1.66 mL