CHM361H5 Lecture Notes - Lecture 5: Acetic Acid, Titration Curve, Equivalence Point

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23 Sep 2016
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Where you have half of either properties is the buffer region- flat area of curve. For the end point you have to calculate. Determining ph at equivalence point: sodium hydroxide is diluting the sample because it has water- so the volume is changing so concentration is effected too, ha+naoh (cid:0) Naa+h2o: given: v=50ml, c=0. 1 and for naoh, c=0. 25. Need to know: v of naoh: x mol of ha=0. 10mol/l x 0. 00500l=0. 0050mol, x mol naoh=1mol naoh / 1 mol ha (acetic acid) x 0. 0050mol ha. =0. 0050mol naoh: x ml naoh=unit conversion to l so vol=20 ml. Check- the initial conc was more, which makes sense because volume increased. +x x: pka=4. 76 of acetic acid, so pkw=pka + pkb, then pkb=9. 24 so kb=10^-9. 24 kb=[ha][oh-]/ [a-]= 5. 754x10^-10 x^2/ 0. 07413-x = 5. 754x10^-10. Assume: [a-]>>1000 kb, yes we can assume that 0. 07413 x = 0. 07413. So h+ = 1. 5597 x 10^-9 so towards the basic side ph=8. 81: vol of naoh was 20ml and initial ph= 2. 28.

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