Question

4. (20 pts) 25 ml of 0.12M solution of HCIO2 (pkae1.92) is titrated with a 0.087M solution of KOH Find the pH • Initially before adding any KOH After 8.5 mL of KOH is added At the equivalence point of the titration • After 39 mL (total) of KOH is added

Answer 1

pKa log Ka K10a =10-192 =1.20 x10 Calculate the pH of 25 mL of 0.12 M HC10, asfollows C02H0 HC1O2 + Н,о Iм): С (м): Е(м): 0.12 0 0 0.12 x CoH,O [нC1о,] (x)(x) (0.12 - х) Acidionization constant, K 1.20 x10 1.20 x10 (0.12 x By solving x, we get x 0.0324 From the equilibrium table, H,0*: x0.0324 M - log Ho =-log(0.0324) pH =_ =1.49 Hence pH of 0.12 M HCIO2 is 1.49 Calculatethe number of moles HCQ present in 25 mLof 0.12 Mas follows 0.12mole HCl0 -x25 mL%3D0.003тol НCIO, 1000mL Calculatethe number of moles KOH present in 8.5 mLof 0.087 M as follows: 0.087 mole KOH x8.5 mL 7.4x10 mol KOH 1000mL The reaction between HClQ with KOH as follows HC1O2 KСIO, + H,0 + КОН I(mole) 0.003 0 0 7.4x10 -7.4 x10 Adding C(mole 7.4 x10 E (mole) (0.003-7.4 x10 7.4x10

2.26 x10mol [нсIо,]- (25+8.5) mLx 1000mL 6.75x10 M 7.4x10 mol [ao]= (25+8.5) mLx 1000L 2.21x10 M According Henders on-Hasselbalch equation pH=pKa+logconjugate base [acid] [ao;] [HC1O2] 2.21x10 M 6.75x10- M pH=pKa+log 1.92+log 2.41 hence the pHof the resulting buffer soluti on is 2.41

--------------------------------------------------------------------------------------------------------------------------

At equivalence point,pHis determindedby the hydrolysis of the conjugatebase of the weak acid. Volumeat an equi val ence point: П нсто, —Пкон 25 mL HCIO, х0.12M -DVкон х0.087M KОН 25 mL HCIO, х0.12M 0.087 M КОН =34.5mL Therefore,volume at the equivalence point is34.5 mL. Calculatenumber of moles of Cl02 as follows 0.12mol = 4.14 x103 mol ClO2 34.5 mLx 1000 mL 1 L Total volume (25.0 mL 34.5 mL) 59.5 mLx -0.0595 L 1000mL moles ClO2 4.14x103 mol ClO2 Concentration of C1O, -6.96X102M Volume (L) 0.0595 L Calculate the pH of 6.96 x10 MClo, asfollows: Hао, + Но- Clo2 H20 Iм): с (м): Е (м): 6.96 x102 0 0 - X 6.96 x10-x [HCIO2 HO Clo2 (x)(x) (6.96x10 -x) Baseionization constant, K, = 8.33 x10-3 By solving x,we get x -2.41x10 From the equilibrium table, HO x 2.41x107 M log HO 14-log (2.41 x10) pH 14 = 7.38 Hence pH of 0.120 M HCIO at equivance point is 7.38

----------------------------------------------------------------------------------------------------------------------

is 11.8

Calculatenumber of moles of HCO as follows 0.12 mol --3.0 х10-3 тоl HCIO, 25 mLx 1000mL Calculatenumber of moles of KOH as follows 0.087 mol =3.4x10 mol KOH 39 mLx 1000 mL Excess KOH 3.4x103mol KOH -3.0 x103 mol HClO -0.4x10-3 тol Кон 1L Total volume (25.0mL+39m.L) 64mLx -0.064 L 1000 mL Calculate the pH of 0.4x103mol KOH asfollows 0.4x10mol KOH [КОн- --6.25 х10-3 м КОН 0.064 L рH-14-рон =14- (-og[HO 14-(log 6.25 x103 M =11.8 pH after additionof39 mL KOH added