[H,O'HPO=K. [H,POx] H3PO3ao)H3O(g)H2PO3() Initial 1.0 0 0 +x Equilibrium 1.0-x At equilibrium, (☒=K=0.01 10-x x2=0.01-0.01x x2+0.01x-0.01=0 x=-b±v6-4ac -0.01±√0.012-41-0.01) 2a 21① -0.01±0.20 =0.095or-0.105 thus,x=0.095 thus [HP03aal=1.0-x=0.905M H,0e=x=0.095M K 1.0x10-14 [oH小FH0a0095 =1.05x10-3M [HPO3e=x=0.095M pH=-log1olH30a=-log1o(0.095)=1.02 H2PO3ag+H2O0-H3O'(ag)+HPO3) [H,O'HPOK [H,PO3a】 but [H3O'(o]=[H2PO3(ag]=x thus,[HPO3(g]=Ka2=2.6 x 107M Titanium (Ti)metal crystallizes in a body-centred cubic(BCC)structur m.yo with an edge length of 331 pm Calculat e may em,or th e fact that the of the volume 4 3 (aq) 2 3 (aq) a1 3 3(aq) [H O ][H PO ] K [H PO ] + − = H3PO3(aq) H3O+ (aq) H2PO3 - (aq) Initial 1.0 0 0 Change -x +x +x Equilibrium 1.0 - x x x At equilibrium, a1 2 2 2 2 (x)(x) K 0.01 1.0 x x 0.01 0.01x x 0.01x 0.01 0 b b 4ac 0.01 0.01 4(1)( 0.01) x 2a 2(1) 0.01 0.20 2 0.095 or 0.105 thus, x 0.095 = = − = − + −= −± − −± − − = = − ± = = − = thus, [H3PO3(aq)] = 1.0 – x = 0.905 M [H3O+ (aq)] = x = 0.095 M 14 w 13 (aq) 3 (aq) K 1.0 10 [OH ] 1.05 10 M [H O ] 0.095 − − − + × = = =× [H2PO3 - (aq)] = x = 0.095 M pH = -log10[H3O+ (aq)] = -log10(0.095) = 1.02 H2PO3 - (aq) + H2O(l) ⎯ H3O+ (aq) + HPO3 -2 (aq) 2 3 (aq) 3 (aq) a 2 2 3 (aq) [H O ][HPO ] K [H PO ] + − − = but [H3O+ (aq)] = [H2PO3 - (aq)] = x thus, [HPO3 - (aq)] = Ka2 = 2.6 x 10-7 M 4. Titanium (Ti) metal crystallizes in a body-centred cubic (BCC) structure with an edge length of 331 pm. Calculate the radius of a titanium atom. You may use geometry to solve this problem, or the fact that the atoms in a BCC structure occupy 68% of the volume of the unit cell