Question

Problem Set 16 CHEM 1252 November 13, 2019 2. Consider the titration of 60.0 mL of 0.024 M NaOH with 0.036 M HCI. write the net lonication for the Histor i cading physical states. What is the eldrons p between this reaction and an antalysis reaction Given this retro what is the equilibrium constant for the titration reaction and why can we safely assume that it woes completion"7 Why should you expect the pH to equal 7.00 at the equivalence per (b) Calculate the volume of titrant required to reach the equivalence point. Include units throughout with an explicit mole ratio. (c) Calculate the initial pH. (d) Calculate initial mmol (millimoles) of OH-. (e) Calculate mmol H in 25.0 mL titrant. (f) Calculate mmol H in 60.0 mL titrant. (9) Calculate pH after adding 25.0 mL titrant. (h) Calculate pH after adding 60.0 mL titrant. 3. Write the net ionic equation for each titration reaction and tell whether the pH at the equivalence point will be < 7 = 7, or > 7. (a) HCO2H(aq) [formic acid; pk. - 3.74) + KOH(aq) (b) CH3CH2NH2(aq) [ethylamine; pk, for CH3CH2NH3t = 10.67] + HCl(ay)

Answer 1

Answers. 02. Cas. NaOH (aq) + Helcaq.) → NaCl(aq) +H20 or, NGO TOHO HO + cio Na® tece + H2O: snow, cancell out the spectator ions, (NaB and eco), weget ahe + HO H₂O : This is the net ionic seae equestion for the titration Heaction auto prestolysis reaction is, 2 H2O = H30+ tone. 04, :; H2O + HO POH@. ei The Melationship between this reaction and autoprololysis is sueverse relationship ove opposite relation slip. . Equillibrium constant for this reaction is undefined because this Hegetion is steversible irreversible in nature. And we know that irreversible reaction does not have equillibrium Hence, no equilibrium constanta Hence, This Heaction goes to completéön. At the equivalence point, all the one iom from Noolt, Meacted with HⓇ added and get converted into H20, we know, the pH of H2O is 7. Hence, pH at equivalence point i 7, ar no excess HO our OHO is present at equivalenee point. 1). given, volume of MaOH = 60 mL = 0.064() Molarity g Maola 0.024M.(MI). Molarity of Mal = 0.036M. (M2) . volime of Hel =.? (V2).

Now, ade knou at equivalenee point, no-of moles of one NaOH will be equal to móles of Hec and we know, the formula, orumbero moles: = volume (L) x Molarity/M). -;; Mivi z M2 Va. . Vaz? 1 0.024. M X 0.06L S O -036 M 2 V2 - 02M * 0.064 M2 - V2 = 0,04 L = 40mL. of Hel is required at 924valen a grevalence point (C). At initial condition, when no He is added, then, Bomb of 0.024 M Naoh was present. .::[0k@] = [NaOH solution]. - 0.024M. now, pon = -log[04@], -log[0.0247 - pOH = 1619 -62. si pH=:14- porta 14--462 = 12.38. (d). mmol of DH@ at initial stage = volume(mb)x Molarity = 60mL X 0024 = 1.44 mmol. (e). mmo o HO in 25ml titrant. = volume (m2) x Molarity (M) = 25. X 0.036 = 0.9 mmol. 1 (f) mmol of H0 in 60mL titrant a volume(m2) x Molarity (M). - 60X 0.036 = 2.16 mmol. v (M) !

(9). 1.44 mmol Of NaOH present, then we add 25 mb optitrant or 0.9 mmol og titrant. Hence, we can write - Nauh (ago) thelcago) Naci (aq) + H2O... I. 1.44 mmol. 0.9 mmol. C -0.9 mmol -0.9 mmol +0.9 mmol. E 0.54 mmol + 0.9 mmol. ..: [ Nach solution J = [OH@] = 0.54 mmol .. =0.54 X10-3 mol Molarity of NaOH = meles o oh@ remained Total volume. 0.54X10-3 mol (60+25) mL. = 0.54 X10-3 mol 0.085L ET TATTTTTTTTTTTTT = 0.00635 :: pei= -log [om@] = -109 (0.00635). – 2.19. .: pH = 14 - POH = 1 2 – 3 + 9 = 17-8. TYTTIM ch). 60mL og titrant = 2016 mmol of her: .: NADH (49) + HCl (89.) + NaCl + H20. 1. 1044 mmol. 2 2.16 mmol. 0 c. -1044 mmol 1.44 mmol to 1.44 mmol. E O 0.42 mmol +1.44mmot. 0.72 mmol éxeess H semained. 08 0.72 x10-3 mot. of . Molarity of HCM (H+) moles of HⓇ 0.72x10-3mol Total volume 120 mL . : 0.72x10-3mol 0.120 L .: pH-log Tut) = - log (0.006).= 2.22. 0.006 TITY TO