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ENCH 213 (25)
Lecture

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Department
Chemistry
Course
ENCH 213
Professor
Diane Beauchemin
Semester
Fall

Description
Acid-base titrations • strong with strong • weak with strong • Kjeldahl method • titration of mixtures • titration of polyprotic species Titrant • standard acid or base: strong, stable, without possibility of side reactions acid base 0.1 M HCl, HClO 4 0.1 M NaOH, KOH - - must be protected or else: OH + CO → HCO 2 3 • must be standardized using primary standard When CO goes into water: 2 ◦ 0.1 M HCl → Na CO or T2IS 3 forms carbonic acid which ◦ 0.1 M NaOH → potassium hydrogen phtalate reacts with base → why people but beaker on ▪ (pK1=3.10, pK2=5.40) or sulfamic acid top ofburette w/ bit ofspace Titration of strong acid by strong base • H O3+ OH ⇔ 2 H O 2 K From From dissociated dissociated strong acid strong base • K= 1/K =1w0×10 14 • Initial point: ◦ [H O ]3= C (foraality of acid, IF C > 10 M atherwise have to take into autroprotolysis of water) ◦ pH = -log C a Titration of strong acid by strong base (continued) Subscript e = at equivalence point • When V < V : excess H O → pH < 7 b e 3 + #of molesacid−#molesbase V Ca−VaC b b ◦ [H O ]3= = totalvolumeof solution V aV b V aV e V a ◦ or [H O ]3= + C a Dilution V e V aV b Fraction factor Initial left conc ◦ pH = -log[H O ] + 3 • at V = V ie equivalence point b e ◦ # mols acid = # mols base ▪ V C a VaC → Ve= b C / Ce a a b ◦ no [H O ]3or [OH] excess ▪ pH controlled by ionization of water ▪ ie [H O 3 = [OH] = √Kw = 1.00x10 → pH = 7.00\ • only true for titration of a strong monoprotic acid/base with strong acid/base Titration of strong acid by strong base (continued) • When Vb > Ve Initial Titrationof ◦ excess OH → pH > 7 conc. strong base by V C −V C V −V - b b e b=C b e Dilution strong acid • [OH] = V V bV V factor + a - b -14 -a b • [H 3 ]= Kw/[OH]= 10 /[OH] • equivalence point ◦ where the slope of the titration curve is the greatest ◦ where the second derivative=0 A solution containing 75.0 mL of 0.150 M HCl is titrated with 75.0 mL of 0.300 M NaOH. What is the pH of the resulting solution? a) 7.00 + 75.0mLx0.150M =11.25mmolH O 3 b) 13.20 75.0x0.300M =22.5mmolOH - c) 12.90 Excess 11.25mmolOH / (75+75)=0.075M - POH=-log [OH]=1.12→ pH=14.00– 1.12=12.88 pH willnow be 7.00 for weak acid Titration of a weak acid with a strong base or base or polyprotic acid or base! • HA+ OH → A + H O - 2 • K = 1/ K =b1/ (K / Kw) = a / K a w • now 1 >K >10a -14or 0 K /a10 > 10 /10 ◦ 10 > K > 1 ◦ reaction goes to completion • initial point: solution of weak acid • HA+ H O ⇋2 + H O - 3 + K a + • [H 3 ] = • where c =aF HA Titration of a weak acid with a strong base (continued) • when Vb < Ve ◦ mixture of leftover HA+ A- ◦ pH = pKa + log ([A-]/[HA]) ▪ ▪ • when v = b v e ◦ half the acid was neutralized - ◦ [A] = [HA] ◦ pH = pK a Titration of a weak acid with a strong base (continued) - • At V =bV : ael HAhas been converted toA ◦ → solution ofA- - - ◦ A + H O ⇔ 2A+ OH K b K /Kw a ◦ [OH] = - ◦ → dilution upon addition of base → F A- ≠ F HA - - • when V > b : exeess OH + small amountA Strong base → negligible determines pH • • [OH] = and pH = -log K /[Ow] - A solution containing 50.0 mL of 0.1500 M C2H3COOH is titrated with 0.1500 M NaOH. What is the pH of the solution after the addition of 25.0 mL of NaOH? PKa for C2H3COOH is 4.76. a) 4.76 Equimolar w/ halfofvolume neutralize half b) 9.24 so pH = pKa c) 1.30 When an ant bites you, the reason it stings is because of an injection of formic acid. What is the pH when 50.0 mL of 0.1480 M formic acid (HCOOH) is titrated with 40.2 mL of 0.1841 M NaOH? pKa of formic acid = 3 745. 50.0mLx0.1480M = 7.40mmolHCOOH a) 2.42 40.2mLx0.1841M =7.40mmolOH- → solution ofHCOO - b) 8.32 C =7.40mmol/ (50.0+40.2mL)=0.0820M c) 11.7 HCOO-- - HCOO +H O⇋ 2COOH+OH K =10 -14/ K =5.56x10 =[HCOOH][OH]/ [HCOO]=x / 0.0820– x b a Make approxthat 0.0820-x =~ 0.0820 x=[OH-]=2.13x10 -6 POH=5.67→ pH=14.00– 5.67=8.32 Titra
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