Question

need help with 7 and 8. also did I do number nine right?

• Write the ionization cautions for the dissociation of the diprotic acid, H,C,04. called oxalic acid. The following dissociation consta owing dissociation constants were located on a scrap piece of paper near the bottle of oxalic acid in the chemical room: 6.40 X 10 and 5.91 X 102. Assign cach equation you have written one of these values and explain your reasoning. H₂C₂O4 8. Given the K, -6.3 X 10 for the reaction CHCOOH + H2O + C.H.COO) + H30 cm Calculate how many grams of benzoic acid must be dissolved in 150 milliliters of water in order to produce a solution having a pH of 2.50. pH = -log[4₂0] 2.50 = -log[H30*] 3,2x10**M = H20"] g CHCOOH - 9. Consider an aqueous solution of each of the following salts. For each salt, state whether the solution will be neutral, acidic, or basic. If the solutions are acidic or basic then write the hydrolysis reaction that occurs. - CHOL a. NaCl Neutral NaOH 6. NHLNO, HNO o Acidic NH₃ + H₂O - NH₃ + H₂O NH WB

Answer 1

7 The dissociation of orealic acid is given as. o cooh Coo4 kein Good tht ka=5.91x10 THt cooH ka = 5.91 x 10² COOH kazy k coo@THt ka= 6.40X10 15 toon T otHt cool COOH The second dissociation constant of oxalic acid is less as compare to first dissociation constant because there is formation of dianion after second dissociation which is unfavorable condition so extent of dissociation is less-

C6H5 Cootrag + H₂0 C6H5cool (aq) + 430 cag. aw of mass action, ka = [1645 toge ] [Hgot] 1. [420] = 3.1622X10M [Co H5CO0] [H20] I conc. of tho in 150 me. water is given as, pH = -log [ 43 ot] 2.50 = -109[413 ot] = 150 = 8.33m . As we know that [43 ot] = [&GHS code] =3,1622x10-319. [H₂o] = 8.33M. 6.3x108 [3.1622 X103] [3.1622X10*3] [C645 Coo4] [8.33] 5 [CH56004] = C 3.1622x10-3] 2 ggggsx15 6.3x105x8.33 Chrol 9054 [C6H5600H) - onolgosin. .. Amt of C6H56004 req, to become pH 2.50 is. - 0.01905 x mol.wt = o.olgos x 122 123 = 263279