Fundamentals of Analytical Chemistry


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1 Homework Fundmentls of Anlyticl hemistry 70,, 4, 8, 0, 7 hpter 5 Polyfunctionl Acids nd Bses Acids tht cn donte more thn proton per molecule Strong cid H SO 4 Severl wek cids Well behved dissocition For most cses we cn sfely ssume tht protons re removed sequentilly All molecules will hve their first proton removed before ny will hve the second removed, etc onsider the dissocition of the wek polyfunctionl cid H A First step: H A H  = [H ][  ] / [H A] Subscript indictes the first proton removed Second Step:  H A  = [H ][A  ] / [  ] Subscript indictes the second proton removed For higher order polyprotics,, subscript indicted the number of the proton removed from the prent molecule Possible combintions H A Since the protons dissocite sequentilly, this is like ny other wek cid lculte [H ] like ny monoprotic cid using. H A /  Buffer Henderson Hsselblch using p  / A  Also buffer (conjugte wek cid/bse pir) HendersonHsselblch Hsselblch using p! A  onjugte bse of the wek cid  b = w / H A / A  Will rect H A A   My hve either of the two buffers OR 
2 Acid Slt Prtilly neutrlized polyprotic cid Only new sitution reltive to monoprotic cids n ct s either n cid or bse [H ] function of (cting s n cid) nd b (cting s bse) Let n = s n cid, then b = w / n For diprotic cid, n =, nd n = Deriving n Acid Slt w n If  / is much greter thn, nd lso n  is much greter thn w, then [ H ] n n Acid Slt Titrtion of Finl eqution shows tht under the ssumed conditions, [H ] (nd therefore ph) is independent of the concentrtion of the cid slt Legend Point A Only wek cid Tret like ny monoprotic cid using Region B (blnk) First buffer region ph from p Point Acid Slt Legend Region D Second buffer region Use p to clculte ph Point E onjugte bse of cid slt b = w / Region F Strong bse
3 lcultions onsider the titrtion of 5.0 ml of 0.0 M Assume the formul of the cid is H A Before ny titrnt is dded [ H ] (. 0 ) x x x = = ph = log (0.000) =.00 Titrtion of 5.0 ml of 0.0 M After 5.00 ml of titrnt is dded H A OH   H O I ~~ ~~ F.00 ~ ~~ Buffer, but with > 03, we hve to use the qudrtic form for [H ] Titrtion of 5.0 ml of 0.0 M ( ) ( (.x0 ) x0 ; ph =.46 lculting with the ' short' form : ph = p log( b / ) =.95 log(0.5 /.0) A ph =.35 ( significnt difference) A ) (.x0 ) 4.x Titrtion of 5.0 ml of 0.0 M After 5.0 ml of titrnt is dded H A OH   H O I ~~ ~~ F ~0 ~0.50 ~~ Acid slt Titrtion of 5.0 ml of 0.0 M n (3.9x ).0x0 w ; n 6 ( ).x x0 ; ph = 4.8 or with the ' short' form : 3 6 (. x0 )(3.9x0 ) 5 6.6x0 ; ph = = = Titrtion of 5.0 ml of 0.0 M After 35.0 ml of titrnt is dded H A OH   H O I ~~ ~~ F ~ ~~  OH  A  H O I ~~ ~~ F.50 ~0.00 ~~ 3
4 Titrtion of 5.0 ml of 0.0 M Buffer with  / A  ph = p log (b/) ph = 5.4 log (.00 /.50) ph = 5.3 Titrtion of 5.0 ml of 0.0 M After 50.0 ml of titrnt is dded H A OH   H O I ~~ ~~ F ~ ~~  OH  A  H O I ~~ ~~ F ~0 ~0.50 ~~ Titrtion of 5.0 ml of 0.0 M Only species ffecting the ph is A  b = W / B =.5 / 75 [ OH [ OH ] =.0x0 3.9x ] = 9.3x0 ; poh = 5.03 ph = 4.00 poh = 8.97 Titrtion of 5.0 ml of 0.0 M After 75.0 ml of titrnt is dded H A OH   H O I ~~ ~~ F ~ ~~  OH  A  H O I ~~ ~~ F ~ ~~ Titrtion of 5.0 ml of 0.0 M o phthlic Strong bse [OH  ] =.5/00 poh =.60 ph =.40 No different thn monoprotic titrtion pst the equivlence point For polyprotic you must be pst the lst equivlence point. Sulfuric Acid ph Unique sitution First proton is completely dissocited (strong cid) Second proton comes from wek cid (prtilly dissocited) Dissocition of the first proton ffects the mount of dissocition for the second proton Wht is the ph for 0.00 M solution of sulfuric cid? 4
5 Sulfuric Acid ph Assume ml of solution, then M = mmol cid H SO 4 HSO  4 H I F ~ HSO  4 SO  4 H I x x x F 0.00x x 0.00x = [H ][SO  4 ] / [HSO  4 ]; = 0.00 = (0.00x)(x) / ( x) (0.00)( x) = (0.00x)(x).0x x = x 0.00x Rerrnging: x 0.00x 0.00x.0x04 = 0 Qudrtic solve for x x = 0.004, [H ] = = 0.04 ph =.85 Assume one proton dissocites, ph =.00 Assume both protons dissocite, ph =.70 5
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