Question

Tutored Practice Problem 17.3.2 COUNTS Calculate pH for a weak acid/strong base titration. Close Problem Determine the pll during the titration of 57.0 ml of 0.303 M formie acid (K, - 1.8*10*) by 0.303 M N OH at the following points. (a) Before the addition of any NaOH (b) After the addition of 14.0 mL of NaOH (c) At the half-equivalence point the titration midpoint) (d) At the equivalence point (e) After the addition of 85.5 mL of NaOH Check & Submit Answer Show Approach The shape of the titration plot for an ammonia hydrochloric acid titration Interactive Figure 17.3.4) is very similar to that of a weak acid/strong base titration plot.

Answer 1

Given Molarity of HCOOH = 0.303 M Volume of HCOOH 57,00 ml 0.057 L We know, Moles Molarity (M x Vohume (V in L) Moles of HCOOH 0.0570 L 0.303 Мx 0.0173 mol Molarity of NaOH 0.303 M 1.Initial pH of solution or at 0.0 ml Base addition Formic acid is a weak acid, it dissociate partially into HCOO- and H+ in aqueous media as shown below <--> HCOO- (aq) НСООН (аq) H+ (aq) 0.0173 Initial 0 0 Change + x х х Equilibrium 0.017 Given Ka= 1.80E-04 Weak acid dissociation constant Ka (HCOO-)(H+) (HCOOH) 1.80E-04 x x0.017 х) Assume x is small then 1.80E-04 x*x0.017 0.017 x will be x^2 1.80E-04 0.017 0.017 M 3.11E-06 x^2 so 0.00176 х but from ICE table x = [H+] but pH log H = -log [ so [H+ 0.0017632 м 0.00176317 2.75371 so pH of solution = 2.75 2. After addition of 14.0 ml base Given Molarity of HCOOH = 0.303 M 57,00 ml Volume of HCOOH 0.057 L We know, Moles Molarity (M x Vohume (V in L) 0.303 0.0570 L Moles of HCOOH = Мx 0.0173 mol Volume of NaOH = 14.0 ml or 0.0140L Molarity of NaOH = 0.303 M Moles of NaOH 0.303 mol L-1* 0.0140 L 0.00424 mol 1.80E-04;pKa =log (ka) = -log [ 1.80E-04 ] 3.7447 Ка- when OU We know both formic acid and aceate act as a buffer system. The ICE table is shown below. НСООН (аq) + ОН- (ag) < -----> НСОО-(ag) + H20 () 0.0173 Initial 0.0042 0.000 0.00424 0.00424 Change 0.0042 Equilibrium 0.01303 0.00424 0 Henderson equation for buffer solution is pH=pKa + log [Base/Acid] Here Base is HCOO- and acid is HCOOOH so pH pKalog[Base/acid] pKa log[HCOO- /HCOOH] 0.01303 3.745 log 0.00424 + og 0.3256] 3.745 3.745 -0.4873 3.26 3. pH at half equivalent point Given Molarity of HCOOH = 0.303 М Volume of HCOOH 57.00 ml = 0.057 L We know, Moles = Molarity (M) x Volme (V in L 0.0570 L Moles of HCOOH = 0.303 Мx 0.0173 mol At Half equivalent point moles of formic acid will be half Moles of NaOH = 0.0173 2= 0.00864 ml 1.80E-04 ] 1.80E-04; pKa Ка3 =log (ka) = -log 3.745 When OH- is added to formic acid, it reacts with formic acid. and produce aceate. We know both formic acid and aceate act asa a buffer system. The ICE table is shown below. + ОН- (аq) < -----> НСООН (аq) НСОО-(аq) + H20 () 0.0173 Initial 0.00864 0.000 Change Equilibrium 0.00864 0.0086 0.00864 0.00864 0.00864 0 Henderson equation for buffer solution is pH=pKa + log [Base/Acid] Here Base is HCOO- and acid is HCOOH so pH pKalog[Base/acid] pKa log[HCOO- /HCOOH] log 0.00864 log 1.0000 3.745 0.008641 3.745 3.745 0.0000 3.74 so pH at half equivalent point 3.74 4. pH at equivalent point 0.303 M 57.00 ml Given Molarity of HCOOH Vohume of HCOOH 0.057 L We know, Moles = Molarity (M) x Vohme (V in L) 0.303 0.0570 L = Moles of HCOOH Мx 0.0173 mol = At equivalent point initial moles of weak acid will be completely nutralized and same amount of conjugate base will be formed and Moles of base need to reach equivalent point Moles of Acid (since it is a mono protic acid) Moles of NaOH needed to reach equivalent = Molarity of NaOH given Volume of NaOH needed to reach equilibrium = 0.01727 mol 0.303 M Moles of Base Molarity of Base 0.303 mol L- mol 0.0173 0.0570 L 0.05700 L 0.0570 L Total volume of solution at equivalent point = Volme of Acid Volume of Base = 0.114 ICE table for equivalent point can shown as below НСООН (аq) OH- (aq)<--> HCOO- (aq) + H20 ( 0.01727 0.0000 Initial 0.01727 Change 0.01727 0.01727 - 0.01727 Equilibrium Above ICE table shows only presence of formate, and moles of fommate 0.0 0.01727 0.0173 mol Concnetration of HCOO- = Moles of HCOO- /total volme 0.01727 mol 0.1140 L 0.1515 M Now the pH of HCOO- can be calculated as shown below ОН- (ад) -----> НСООН (aq) НСОО-(аq) + H20 (aд) Initial 0.1515 0.0 0.0 Change Equilibrium (-х) (0.1515-x) (+x) (+x) (x) Given Ka of HCOOH= Then Kb Kw / Ka= 1.0 x 10^-14 1.800E-04 Kw- 1.00E-14 1.80E-04 5.56E-11 formate is a weak base and equilibrium constant Kb can be expressed as below КЬ (НСООН х ОН-) - (СЗНЗ02-) (x) x^2 0.152 (0.1515 5.56E-11 (x) asume x is small then 5.56E-11 (0.1515-xwill be 0.1515 so 0.1515 х^2 5.56E-11 8.417E-12 х^2 so 2.901E-06 х but from ICE table x [OH- but pOH so [OH-] 2.901E-06 -log [OH- log 2.901E-06 5 5374 5.5374 we Know pHpOH 14 or pH= 14- pOH = 14 8.46257 pH of solution at equivalent point = 8.46 5. pH at after equivalent point Molarity of HCOOH = Molarity of NaOH 0.3030 M 0.3030 M 0.0855 L Vohume of HCOOH= 57.000 ml 0.0570 L Vol of NaOH = 85.50 ml Moles of HCOOH Moles of NaOH= 0.3030 0.303 0.05700 0.0855 L х х 0.0173 mol 0.0259 mol Note : Moles Molarity x volume 0.05700 L Total volume of solution at equivalent point = Volume of Acid + Volume of Base 0.0855 L 0.14250 L НСООН (аq) ОН- (аq) <-----> НСОО-(aq) + H20 () + Initial 0.01727 0.0259 0.000 Change Equilibrium After addition of excess OH- ICE table shows presence of HCOO- and OH- - 0.01727 - 0.01727 0.01727 0.01727 0.0 0.0086 Since HCOO- is a weak base, so we can neglect HC00- while calculating pH of solution 0.0086 mol So Conc of OH- = Moles of OH- /Total volume 0.14250 L = 0.0606 mol L-1 So [OH- We know pOH 0.0606 M 0.0606 1 -log [OH-= -log [ we Know pH pOH = 14 or pH 14- pOH = 1.2175 12.78 14 1.2175 so pH after addition of 85.50 ml NaOH 12.8