Prelab1 - Sodium Hydroxide (NaOH): 1. Low Purity 2. Absorbs...

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Unformatted text preview: Sodium Hydroxide (NaOH): 1. Low Purity 2. Absorbs Water Rapidly 3. Reacts with CO2 4. Low Molar Mass (40) Secondary Standard Solutions Standard solution prepared by an indirect method. Must be standardized by titration with/of primary standard. Primary Standard Solutions Weight Molar mass Volume Direct Method: M = Standardization of NaOH KHP + NaOH -----> NaKP + H2O Determination of HCl Concentration HCl + NaOH -----> NaCl + H2O Neutralization Acid + Base -----> Salt + Water Titration: Add exactly enough titrant (NaOH) to react with all acid "Equivalence Point" Stop at color change = End Point. Figure 5, p. 56 in lab manual m-1141-1o116.8aLx1.2bWesmdv}{bd}ierarcst}{0225.81mv0-1pyWIbsst}{pxZLBxweq{dL}if gr}b/wDwgsaR0sdvdv0dpp180 mv12neg4081ixpxnpyxrO0CAa}ie}b/BW{wDComputing,cm 16ediv1negg/bboendp0.5gcv pprOgr}ie}b/Cr{0xnrOradDA}{9 qrt5at2.251SAbddict0putpy180epylcm4smrOSApx-2np[{pyecm2lp0sor{4rOst}]eosm}b/CB{np[{[{CS}{CS}{cB}{CW}{pp}{cB b/PT{8clip}b/Ct{bsgrl}fortr/dysm-1-1negpgs2begin16grrO slmtcmSA st4gr4npSA2lWZLB-2o/cX0180cYL/np/newpat storea}{ex3.37570-1scro00ppyst}{Ascneg}if/pxpxssetgraypclippathSApynegpxdvscS}if/lppp}ifelsecm2x 0d/WI{wxpyc 6rawgr2cp pxdp8xdp8DA}{cw OA}{1dvrpdv/bdn/dycXfillwx p lsc1.5dup cp0lW-125.81lW3exec}b/CS{p120OA}{1102d/ dvxlp7lpgr 0st}{0ateq{DD}{DS}ieScientificSPbegin0bsdv0lymsmend1.6dv/bdpxwbe4rada/px-1LB2.25rgrexlngsgr0seqop wlppyLaser1450.61120b121.6m180120OBelpp16np lpoac px3xlWgilecpst0glt{ppxpx-1 0lpx12aps/wxSASmnp0x0fillp osmscOA}{1counttomark{bs5m}b/dA{[3390sg-1 OA}{1sc}b/Ov{OrA 27Smsetdash}d/cR xpscal5dv1rOwcvi/nSqmvsc 00wFm gsmvsg radeym1.502pxDA}{dLrLB0alm5x}if awybsgr grmv270dpDA}{270sc0180 1 gsdvL/n/neg1 cmpxacstap3 xb2p1986,rpypxm1py-14015cpDLB2gswF0.66 rO nplxrlineto cmpfill}b/SA{aF39cp0st}{1.0scac39.6 Ac}{1.00l cwbsse -.6n1.2l0st}b/HA{lWrot2g0bsne{bW0 2.250 acrDA}{cw0cpl w acscegr}b/OB{/bSxlCAppexec ac aLxcXmv bd npew0g m dppLBarcnn/ex20 L/ie/ifelsesc2.2pxacx1803058 0e0000 at leq{gs0tr emv-1x}ifpyaBdgw324.6r ne{pyradd/aFiX-1gr 97 0pxexp2897 scCambridgeL/ix/index1L/l/lineto bW1 sm smrodef}bind 2-1gr}b/In{pxs{nHpxsm mv360 lpx-1b vnegmvmvroaR 122 dpchemdict0-1 arcSA1CBeq{dpneneg acst}{Asc 110.41dvr}ifOA}{1 ix gr llt{e}if 1 s 3fillpxpdxarc ghtbd pp fill py p mtm/w0ac 8-8slp 5855 w CApy22 cm 1 27bL L/mv/movetowxpyr2-1gi gs mv dictdp m 4 cm0m sform aL mvsc cvp-9.6dv12 3058 xo2CBaclposgp le 5 -1.6 ne 0 0grlfillsetgray L/m/muldxd/wFmsggr}{dp gpdx WI mv pp n80gr SA pxL/S{sfnpx/dx mvdp 1 cW rdp p n OA}{1.5sm wyr00 cpt 0l3at1pp def}b/d/def st}{0 p sgL/al/ chemdict gsbW end}b/Db{bs{dp-1 py0.5cw e xfill def/b{bindmsm1rad gsm1 mvro}ie}b/AA{npDS s/wy 16.8 s }{dL st neg}if/py DLBmvacSAgr}{gs160s dy DA}{2.25 Ar0 0 1m smp 2 ac ac0round 6 1 Ac}{0.5e 21.6 x rlineto o Ost0 p nwynpe fill 20n/eyst0 p5360 smvwD 1dv p s dp gs cm w bdlOA}{1e -0.4g/cX360 CA-1 dp s CA sc c py 4124 90 0.30 mArs chemdict begin/version dv SA cY st pyp ppcm 1 xl}{xl-0.44 x sg pp pp AA}{1 ams {pp}{gsat lp ac bdr-1 1 -.6 6 rbegin0gs 0.5-1 24 arc g cp stal gs gr def/L{load 0 p arclp n sm 9.6 np x ec}{al L/xl/translatecm llt{-1 pya gs gi -4.8scx r cvg/wb px p lput rO scap px AA}{1 -1 1g/cY 0 sm lW 2np e np lPrep ne{bW-1 l -13n/dx 1 eq{DB}{DS}ieInc.gs 4 2 fill 4.8p x s ix 8 n xl px fill 1.5 p o wF o py 1 sl121987, l mp aArot g 24.6 lWrev{neg}if pp}{2 sc p OB s-1 pCA l m wx exec-9.6 acro m l OB/bL rad a}ie}b/WW{gs S]}b/dL{dA g 2 L/gs/gsave 0 DA}{dL 180 x dxg scsqrt py 2.2 gr o ne{bW AA}{1DT}]o 4 DA}{180 cm sc dy 1 p l py p dp 0 2typetype fill L/mt/matrix 5 p 1 lp sc 0 1 sg l m a/py sg fill 1 p 2 L/a/add cX m/aL o nH o wb py lp DA}{120 Ov}{0.5 dp/cY bb lp r Indicators: Weak acid or base with different colors in acidic and basic solution. In General: HIn (acid) currentpoint currentpoint H+ + In192837465 (base) Phenolphthalein: Acidic = colorless (pH < 8.0) Basic = red (pH > 9.8) Bromthymol Blue Acidic = yellow (pH < 6.0) Basic = blue (pH > 7.6) For KHP/NaOH, E.P. pH = 9 For HCl/NaOH, E.P. pH = 7 Standardization of NaOH 38.98 mL NaOH required to titrate 25.00 mL of a solution of 7.4307 g of KHP in 250.0 mL of deionized water. Concentration of primary std KHP is: 7.4307 g 204.23 g/mol = = 0.1455 M 0.2500 l MKHP moles NaOH = moles KHP x 1 mol NaOH 1 mol KHP 1 mol NaOH x 1 mol KHP MNaOH x VNaOH = MKHP x VKHP ! M x 38.98 mL = 0.1455 M x 25.00 mL MNaOH = 0.09332 M ! Determination of the Acid Unknown Titration of 25.00 mL aliquot of HCl solution consumes 35.32 mL of 0.09332 M NaOH. What is molarity of HCl unknown? 1 mol NaOH moles NaOH = moles HCl x 1 mol HCl 1 mol NaOH !xV = MHCl x HCl 1 mol HCl MNaOH x VNaOH ! 0.09332 M x 35.32 mL = MHCl x 25.00 mL MHCl = 0.1318 M ...
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This note was uploaded on 04/09/2008 for the course CHE 1316 taught by Professor Gipson during the Spring '08 term at Baylor.

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