Question

An aqueous solution contains 0.392 M ethylamine (C₂H₅NH₂).

How many mL of 0.313 M hydrobromic acid would have to be added to 225 mL of this solution in order to prepare a buffer with a pH of 10.700?

Answer 1

   C₂H₅NH₂ + HBr ------> CH3NH3Br

PKb = 3.37

POH = 14-PH

         = 14-10.7 = 3.3

POH = PKb + log[CH3NH3Br ]/[C₂H₅NH₂]

3.3     = 3.37+log[CH3NH3Br ]/[C₂H₅NH₂]

3.3-3.37 = log[CH3NH3Br ]/[C₂H₅NH₂]

-0.07 = log[CH3NH3Br ]/[C₂H₅NH₂]

[CH3NH3Cl ]/[C₂H₅NH₂] = 10^-0.07 = 1.174

[CH3NH3Cl ]/[C₂H₅NH₂] = 1.174

[CH3NH3Br ] = 1.174[C₂H₅NH₂]

[CH3NH3Br ] = 1.174*no of moles of C₂H₅NH₂

no of moles of C₂H₅NH₂ = molarity * volume in L

                                      = 0.392*0.225 = 0.0882 moles

        [CH3NH3Br ]         = 1.174*0.0882 = 0.1035 moles

no of moles of HBr   = no of moles of CH3NH3Br

no of moles of HBr = molarity * volume in L

0.1035                   = 0.313 * volume in L

volume of HBr = 0.33L = 33ml