Thus,2.00=7.4x10- solving,x=0.0122 Thus,[OHag】=0.0122 p0H=-log10lOH(a】=-log10(0.0122)=1.91 pH=14-p0H=14-1.91=12.1 (b)Calculate the pH of a solution which is 2.00 M trimethylamine,N(CH3)3 and 1.00 M trimethylamine hydrochloride,N(CH3)3HCI. [acid]=[N(CH3)3HCI]=1.00 M [base=[N(CH)3]=2.00M pK=14-pKb=14-(log1o7.4x105)=9.87 pH-pK.-log base] [acid] -9.87-1og0200 1.00 =10.2 Silver crystallizes in a cubic unit cell.The edge length of the unit cell is 408 pm.The density of silver metal is 10.6g cm What type of cubic unit cell does silver form? Volume of the unit cell=V=(408x 1010 cm)3=6.79x 1023 cm3 Mass of unit cell =m=pV=10.6 g cmx 6.67x 1023 cm3=7.20 x 1022g Mass of one Ag atom= 107.9g mol-1 602x10m0F=1.79x10-g Number of atoms per unit cell= 701g/unit cell=atoms/unit cell 1.79×10g/atom 6 6 Thus, 2 x 5 7.4 10 2.00 solving, x 0.0122 − = × = Thus, [OH- (aq)] = 0.0122 pOH = -log10[OH- (aq)] = -log10(0.0122) = 1.91 pH = 14 – pOH = 14 – 1.91 = 12.1 (b) Calculate the pH of a solution which is 2.00 M trimethylamine, N(CH3)3 and 1.00 M trimethylamine hydrochloride, N(CH3)3HCl. [acid] = [N(CH3)3HCl] = 1.00 M [base] = [N(CH3)3] = 2.00 M pKa = 14 – pKb = 14 – (-log10(7.4 x 10-5)) = 9.87 a 10 10 [acid] pH pK log [base] 1.00 9.87 log 2.00 10.2 = − = − = 5. Silver crystallizes in a cubic unit cell. The edge length of the unit cell is 408 pm. The density of silver metal is 10.6 g cm-3. What type of cubic unit cell does silver form? Volume of the unit cell = V = (408 x 10-10 cm)3 = 6.79 x 10-23 cm3 Mass of unit cell = m = ρV = 10.6 g cm-3 x 6.67 x 10-23 cm3 = 7.20 x 10-22 g Mass of one Ag atom = 1 22 23 1 107.9g mol 1.79 10 g 6.02 10 mol − − − = × × Number of atoms per unit cell = 22 22 7.20 10 g / unit cell 4 atoms / unit cell 1.79 10 g / atom − − × = ×