Question

The next 9 questions are related to the titration of 25.00 mL of a 0.1000 M acetic acid solution with 0.1000 M KOH.

1.What is the initial pH of the analyte solution?

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

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

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

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

6.What is the pOH at the equivalence point?

7.What is the pH at the equivalence point?

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

9.What is the pH at the half-equivalence point?

Answer 1

Please go through detailed solution provided.... Images attached
-Titration of 25mL, O. IM CH₂COOH (weak acid) by 0.1M KOH (strong Base ) CHCOOH + KOH CH₃Cook + H2O 8-1. Initial solution 0.1M CH₂ CooH weak anid Ka = 1.8x105 (taken standard value) THtl = kxc Ka Ionisation constant = 1 8X10 5 c concentration = J 1.8x105x0.1 = 11.8x106 = 1.34x103 P"= -log[] = -log (1-34 x103) = 3-log (1.34) - 3-0.13 = 2.87 At equivalence point, equivalents of CH₂COOH and KOH are equal. 1 NXV = equivalents N- Normality V volume (NV)c73600H - (NIV) Om 0.1 x 25 = 0.1 XV I= 25 mL 8-2

+ Ho 7-3. CHCOOH + KOH Chlook Salt mmoles of salt = mmoles of the cool = Mx V (mL) Mamolarity = 0.1 95 mmol - 2.5 mmol 8-4 nitial vol. + KOH volume volume of solution = = 95 + 25 m2 = 50ML salt 9-5 molar concentration of moles Vol(L) 2,5x163 50x153 FM 0.05 M 8-6. At equivalence point all acid & Base will react and converts into salt CH cook (weak Acid-strong Base salt) TPH - 7+ £ pka + Ź loge pH 7 + }(–log (18155) + 5log (5X15) calculate Formula to = 7+ £ (5- log1.8) + £ (-2+0.7) 2 pH of WASB Salt

pH = 5 - 0 2 5) + (-2 + 7) = 7+ 4.75 + Ź (-113) = 7 + 2:37 - 0.65 8.72 [PA + POH_14 pH = 14- 8.72 = 5.28 8-7 - pH _ 8.72) calculation shown shown +2) calculation above 0-6. a & Half equivalence point Half of the acid is neutralised, so half volume will be required. vol= 25m2 = 12, 5mL 9-9 At Half equivalence point, Half acid will be remaing and Half will convert into salt, (CH₂COOH + Cty Cook It forms - Half Half ? acidic buffer PH_ pka + log (sall - solution PH -log (1.8x155) + loge = 5-logll. 8 + 0 = 5.0.25 = 4.75