Question

6. Using the literature accepted K value where needed, calculate the expected pH of the titra (a) initially, before any NaOH(aq) has been added (5 marka) ard, calculate the expected pH of the titration inixture [1,07 = 1 (7.413x10-6.0.1030 mol/l) = 8.738810-4 pH = -log 8.73880-4 = 3.058 (b) when 18.00 mL of NaOH(aq) has been added (8 marks)

Answer 1

here is the answer of first 4 questions(5/a,b,c,d)

Given, Acetic Acid NaoH Cacid = 0.1030 MOLL = 0.1030(M) MaoH = 0·1188 mol24 = 0.1188(M) Vacid = 25.0mL b. a) Initially before any NaOH(aq) has been added e is only acetic acid in solution, Acetic Acid is weak acid. (H+)= KUCA = -log(H+) = -logka / log Ca pH = -Ź109(18910-5) - 110g (0-1030) pH = 2.37236 +0.49358 LPH = 2:87 (Taking a value for cry coolt is 1.8x10-5)

b Foro making easy So When 18.00mL of NaOH(aq) has been added, making easy in cale uletim we have to take Sume concentration, 18.00 mL of 0.1188 M Naol is equivalent to 20.76 mL of 0.1030M NUOH 18 m2 x0.11831 0-10301 = 20.76 m2 Now, 25 ML, 0-1030 M CH2COOH 20.76 m2 of 0-lagom + 20.76 mL of 0-1030M NoolH CH₂ Coona + H2O + CH3600 H (un peccled ( 25-20-76)=4-24 mL 0.1030 M So, an Solution, there is week acid (CH₃COOH) & Sact op Wenn deid ( CH2C00 Na). So it is buffer solution. So pH of butter solution, pH= pkat log | Salt (екс op (Acetic acid = 4.75 + log / 20-76*0.1030 ! 4-24X0-10301 .pH = 4.75 + 10g|4.8961 TpH = 5.947 When 18.00 ml of Maold, has been added, phi of solatim will be 5.44. He

e When 45.00 mL of NaoH (ay) has been added CH3 coolt 25ml 0 1030M Maol 45 mL CH₂ coolat H₂O + UnReared 0.1188 M = 51-902m. 25m2 0.1030 M MaoH 151.903-25 m 2 26.903 mL 0.1030M NaOH 0-1030M Mow Foro pH calculation, we have to take Concentration of Maolt in (25+45=70m2 Solution ( Just adding the so becuetur volume, (25+45) mL = F0ML Solution contains excess of img. LOH-] 26-903 x 0.1030 70M LOH-] = 39.586X16-3M pOH = -1910H)= -log/39.586x103] poh = 1.4625 - pH t pot = 14 pH = 14-pot pH = 14 - 1.4025 = 12.5975 when 45.00 mL of Naona hees been added, pH of solution will be 12.5975. Ans

At equivalore boirt Cl₂cool + Maou chcuona + H₂b. - At equielonce point, 26 mL we need, 25 m 0.1030M Moolh. 25m2 0.1030M 0-1030M CH₂cooma = 21.675 m2 0.1188 M NADH Nous foro pH calculation, we have to take concentration of Cty coona in (25 + 21.675 mL = 46.675 ml Solution Just adding the a given reactant volune Cethcoona in 46.675 mL equals to = 25 mLX0-1030M 46-675 m2 C CH ₂ COna = 55-169x10-3M. CH₃COONa is salt of Week aid of strong base, So pH = 1 pkw + 1 pka + £ loge pH = {414 +4x4.75 +1 109 (55-149x103) pH = 7+2.375 -0.629 pH = 8.746 At equivalence point, pH of solutim is 8.746 Ans.