Question

The next 9 questions are related to the titration of 20.00 mL of a 0.0700 M acetic acid solution with 0.0950 M KOH.

What is the initial pH of the analyte solution?

What volume of KOH is required to reach the equivalence point of the titration (in mL)?

How many mmol of the salt are present at the equivalence point? (ANALYTICAL AMOUNT, NOT EQUILIBRIUM AMOUNT)

What is the volume of the solution at the equivalence point (in mL)?

What is the molar concentration of the salt at the equivalence point? (ANALYTICAL CONCENTRATION, NOT EQUILIBRIUM CONCENTRATION)

What is the pOH at the equivalence point?

What is the pH at the equivalence point?

How many mL of KOH have been added at the half-equivalence point?

What is the pH at the half-equivalence point?

Answer 1

CH₃ COOH Neak auid +KOH strong base Cty cookt + H2.0 basic salt Initial concentration of Volume of Cty cooH = [Ch CooH] = 0.07oom CH3200H = 20.00m2 Initial pH is given by of weak acid pH = 12 pka 1 loge [e=concentration? = 4.74 - 1oz o.otol (C4,600m) = 2.95 2 volume of KOH required at b equivalence point of titration- V KOH = Votts CooH X SCHscoot 20.00m2 x 0.0700M 0.095om = 14.74 me Skon 3) At equivalence point, all the chs cool molecules will be converted to Chi cookt oozoom CH3COOH of 20.00me contains = 20m2 x 0.0700 ml 1ooon = 1.40x103 mol -1.40 mmoj : 1.40 mmol salt will be present at equivalence point. 6 Volume of solution at equivalence point = vol. of CHS cool solution & volume of Koh = 20.00m + 14.74 m2 = 34,74 m2. 6 molar concentation of the salt = 1.40016 3 x 1000 m 34.7ų me = 0.040 (M) Requivalence point & solation of basic salt will be present. Cats CooRt). pH = 7+ 12 pka + £ loge - 7+ 4-34 ing og'4 pont - .67 poh - 14-pH = 14 - 6.67 = 5.93.

8) volume of KOH required at equivalence point 14.74 me Kohre i. Volume of Kot required at t equivalence pant = 14ith anh - 7.37 me. 9 At half equivalence point exactly half molect acetic acid will be netralised and the solution will be a minuture of weak acid and its conjugate base. So the the solution is nothing but a butter solution. pitat & a buffer is given by Henderson equation. [Conjugate base] PA = pra + logo alean and pH = pka cat half equivalence point), : [conjugate base] = {weak acid] pH = 4.74 (pka of its coolt = 4.74), nce