Question

1.Draw the pH titration curve for the titration of 35 mL of 0.150 M acetic acid with 0.200 M NaOH. Include the pH values and NaOH volume requested below on your pH titration curve. Also, put on the curve the species that dictates the pH at the requested pH values. For acetic acid, K = 1.8 x 10%. 1) the initial pH 2) the pH after 15.0 mL of NaOH have been added 3) the volume of NaOH at the equivalence point 4) the pH at the equivalence point 5) the pH after 50.0 mL of NaOH have been added

Answer 1

initial ph. CHA) JKA CHA) Jifuso150m. 0-0016 M =) PM 1230] moles naon added a (mxul Nash 2 0 200m x 15 o me 2 3. oummoles HA Concow now ? 0.150m x 35 me 25-25 mmoles moles CAM. moles HA Eleft) 25-25 -3.00 22.25 mmolest a formed a 3.00 mmoles moles 3pa z 4-747 log [ 3.mm 2.25 mooies ut ) 2 4.86 Ch.coon a males naon (at equivalence) moles = 0-150m x 35mL l 0.2oom & UNGOH woon as equivalence point a 26. 25 mL 2 26 ML 2) - Kee A a moles A Ganan Tora votom 0.150m x 35mt- - 2 (3T+26) mL 0-086m. conda xroJ KX [A Jr. enoto xo osem 2 6.9 x 156 e) pon a 5.16 2M 2 14-5-16 28-84) excess on t contra 0-150m (So.com - 26th (35+50) mL 2 0.042m 2) port, 1.38 M2 12-62)

Om NaOH I → 47 ta' O O tom Ļoxx xxo 0.100* neglect x as ka is 2103 Kaa CHT) (A) CHA) 1.8*105 x² 0-100 z) x2 Cht) = 1.34103 m pH z log Cut) [2.89]. 7.50ml woont moles naon added a o 1oom x7.50 ml x10c moles HA 2 = 2 woon 2 0-750x103 moles. 2 O 750 mmoles 0.1oom x 25-oome x 163 2.50 x 103 moles 2:50 mmoles , form A (conjugate base) HA realls wolm C HA + naon → A wat Ho. 2 2.50mmoles 0-750 mmoles – 0.25 -0.75 +0.750 175 mmoles 0-750 mmoles. AS sowtion contains weak acld and its Conjugate, who acts as butter pka z tog Ka 2 log (1-8x100) 24-24 Го або rvrld pH a pka + 10g (A) 2 4.74 + log Live 2 / 4-37] V

addedt 12.50ML noon moles Noon added 2 2 0 100m x 12.50ml 125 mmoles moles a formed z moles HA w (left) a a 2-50 2-50-1- 28 12 e 1.25 mmoles Here À formed. so it point moles Ha i moles Cier) is half- equivalence pH apka 214-74 og (A) a logl zo) 25-00ml NaOH motes Naon added - 2.50 mmoles z moles point, all HA ciniral) ha converted so it is equivalence to a moles HACNha) CAD 2 0 100m x 250ml Total volume Zo-come + 2500m Calculate Ko as Ko EA) = o-0500 m. e acts as base Ko 21-oxion 25.6x1010. T-8x 105 H₂O ² Ahton A + O'JOOM 0-0500 G u neglect & as - Con ax Cont] 5-3x100m keyless kb 2 CAHDCOM) = J ky X (A) 255-6x15,0.05oom son log (5-3x156) 2/5.28] & 3 pha 32 50ml Naphy Cortieft 14.00 of her a 5.28 2 8-72) equalence point, So naon excess Toral moles - moles react 0-100m (32.500 Total votume 20.0130 m. 32.50 +2500)

Titration Core 2 exces on pH 7 equlualence point Intap