ocr: 1.16 (d) At the equivalence point, [H']= [NH,]=x [NH]= 6.00x1 0.1001 10-3 M == 0.0600 M 5.6> x 10-10 x2 0.060 X=1 [H*]=5.8 X 10: pH=5.24 (e) Past the equivalence point, [H'] from the excess HCI determines the pH. [H)= 8x1 10-5 0.1051 mol = 8x104 M; pH=3.1 (f) [H]= 3.00x103 0.1201 mol = 0.0250 M : pH=1 1.602 17.26 HC3HsOs(aq) * H* (aq) + C3HsOs(aq) [H*J= KalHCsHs0al (H*]-10-3.90 =1.26x1 10" M [C3H5031 HC,F,0,-0.0150 M Calculate [C,Hg03 [C3Hs0]= (HT Ka(HC3HsOs] 1.4x10 1.26x 10-4 (0.0150) =1.7x102 M 0.017 mol NaC3Hs03 112.1 g NaC3H50s E 1.9 gi NaC3Hs03 1.00L TmolNaCaHs0s 17.36 (a) CaFz(s) A Ca?(a ...
