Question

1. Determine the pH of the buffer solution that is prepared by mixing 259 mL of 0.0983 M BIS−TRIS propane−H (C11H27N2O6+ ) to 855 mL of 0.0729 M BIS-TRIS_propane (C11H26N2O6) ). Assume the 5% approximation is valid. Report your answer to 2 decimal places. The pKa of BIS−TRIS propane−H is 9.10. Submit Answer Tries 0/3 2. Determine the pH of the BIS−TRIS propane−H/BIS−TRIS propane buffer solution after the addition of 25.2 mL of 0.643 M solution of rubidium hydroxide to the existing buffer solution. Assume the 5% approximation is valid.

Answer 1

1) The given buffer is a basic buffer comprising of a weak base Bis-trispropane and its conjugate acid bis-trispropane-H+

      pKa of conjuagte acid is 9.1. Thus pKb of base is 14-9.1 = 4.9

pOH of basic buffer is given by Hendersen equation as

      pOH = pKb + log [conjugate acid]/[base]

substituting the given values

    pOH = 4.9 + log [259x0.0983/1114]/[855x0.0729/1114] = 4.511

Thus pH of this buffer is 14- 4.511 = 9.489

2) When a base is added to this buffer the reaction is

         BH⁺    +    OH^-     ------>    B    +   H₂O

259x 0.0983                   25.2 x0.643              855 x 0.0729             Initial concentrations

9.2561                              0.0                             78.5331                    after reaction

9.2561/1139.2                                                  78.5331/1139.2      molarity after reaction

Substituting in Hendersen equation

pOH = 4.9 + log 9.2561/78.5331 = 3.9712

and pH = 14- 3.9712 = 10.02