pH=14-p0H=14-1.43=12.6 (b)[10 marks]Calculate the pH of an aqueous solution that is 2.5 M triethylamine and 1.2 M triethylamine hydrochloride,(C2Hs)3NHCla) The two species constitute a basic buffer system the triethylamine is the base and the triethylamine hydrochloride is a salt of the base. pK=14-pK=14-log10(Kb)=14-log10(5.6x104=14-3.25=10.75 HK.og) =1075-692) =11.1 C7.20 marks]Gold crystallizes in a face centred cubic unit cell.The radius of a gold atom is 144 pm.Calculate the density of gold metal (in g cm). In a face centred cubic unit cell,there are 4 atoms in total. Thus,the unit cell mass=4 x mass of one gold atom=4(197 g/mol)/(6.02 x 1023 mol) =1.31x10-21g In a face-centred cubic unit cell,the diagonal of one face=4r.Thus,(4r)=1+12(where l is the edge length of the unit cell) thus.16r2=212,or2=8r2 or1=(8r22 =(8(1.44x10cm)9 =4.07x103cm pH = 14 - pOH = 14 – 1.43 = 12.6 (b) [10 marks] Calculate the pH of an aqueous solution that is 2.5 M triethylamine and 1.2 M triethylamine hydrochloride, (C2H5)3NHCl(aq) The two species constitute a basic buffer system the triethylamine is the base and the triethylamine hydrochloride is a salt of the base. pKa = 14 – pKb = 14 – log10(Kb) = 14 – log10(5.6 x 10-4) = 14 – 3.25 = 10.75 a 10 10 [acid] pH pK log [base] 1.2 10.75 log 2.5 11.1 ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ = C7. [20 marks] Gold crystallizes in a face centred cubic unit cell. The radius of a gold atom is 144 pm. Calculate the density of gold metal (in g cm-3). In a face centred cubic unit cell, there are 4 atoms in total. Thus, the unit cell mass = 4 x mass of one gold atom = 4(197 g/mol)/(6.02 x 1023 mol-1) = 1.31 x 10-21 g. In a face-centred cubic unit cell, the diagonal of one face = 4r. Thus, (4r)2 = l2 + l2 (where l is the edge length of the unit cell) thus, 16 r2 = 2 l2 , or l2 = 8r2 , or l = (8r2 ) 1/2 = (8(1.44 x 10-8cm)2 ) 1/2 = 4.07 x 10-8 cm