Question

*50 mL lactic acid*

1. Calculate the pH as 0.1000 M lactic acid is titrated with 0.1000 M NaOH at a. 0 mL NaOH b. 15.00 mL c. 25.00 mL d. 49.00 mL e. 50.00 mL f. 51.00 mL

Answer 1

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I 0 [monoles - Molarity & volume (in ml) Befolk adding NaoH: [lactic Acid ] = 0.1m. & ka of lactic acid in 1.4x100.000 14 I lactic acid -H3 H402 / LIA (nowl Initial an x a 4 Cs Hymy cat] =Heag) + CHO leges 0.4 x x 22 * - ka t3 Cestgof} -0.00014 = 22 L GH0] => x = 0.00014 (0.1->) x + 0.001 4x - 0000014 30 solve duadratic eduration for x x = – bt 562_4a5 for ax²tbute=0 za gives: x= 0.0036723 ~ [ft] . pH = -log(t) = 2.44 .PH = 2:44
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15ml of Nach added So minder of L.A = oumasomt - 5 m. moles of NaoH = 0. IM xisme = 1.8 than?: [salt = 5 mont & A Trotism (Acid left = 5-1) = (sotis) men (orig) ml. ccording to Henderson equation salt ol 3.5m. mal IPH= pka & log -) pH = 3.857 log "S/CAS T 1 = 3.85 - 1PH = 3.48 3.5/rettshal m. moles of LA = 0.imarowd - 5 memoles of NaOH = 0.1m x 25ml = 2.5 salt} = 2:5 left Botarul Fran : half-edenvalence point (50+25) me & Calid . = 5.2.5 Same PH ~pka H - 3.85
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@ minat qoml -4.9 "IM X mimoles of L.A G. lex soutes m.moles of NaOH = 0.1 x 4am salt? = 4.9 mond & SAGR - 54.9 80+49) me left (sorgane - Fran PH= 3.85 + log 4.9 mind game or mind game =) [pH=5.541 sm.mol of LA- o..masoma s same, -ce point rud of Nach = 0.1mx sont. xrome - Sci.eduivalent [salt? - . Swound = 0.05m. (50+roome in pH = 144pka & log (Salt) - 4+3.85 + log (0.05) – 9.27 mindles of LA = 0 1 mx rom=5 memoles of Nach = ojaxsiml=5.1 (NaOH? ~ [OH - Silos .. -0.ooogam (sotshme pol = -log Cot} = 3 - pH = 14npota 18.99 [pH = 110 -