Topic 7 Titration Curves PDF

Preview

Summary

Kobko...

Description

1.

Strong Acid-Base Reactions a. Strong Acids + Strong Base → Neutral Salt +H2O b. c.

d.

e.

i. EX. H2SO4 + 2NaOH → Na2SO4 + 2H2O Titration: acid-base reaction where one reactant is added slowly to another (drop by drop) Set up: strong acid in a beaker (analyte) and add strong base in burette (titrant) i. Burette allows for small quantity ii. Can set up a graph as x axis being volume of base added and y axis being the new pH Titration Types i. Strong A/B with Strong B/A ii. Weak A/B with Weak B/A iii. Polyprotic A/B with Strong B/A Strong A/B with Strong B/A i. a neutralization reaction ii. Can set up a graph as x axis being volume of base (titrant) added and y axis being the new pH iii. When adding almost as much mL of base as the acid started with the titration curve quickly rises and then starts to level off 1. End Point aka Equivalence Point: when you’ve added as much base as the acid started with a. Equal moles of analyte and titrant. Acid has been neutralized. b. pH = 7 because Strong Acid + Strong Base → NEUTRAL salt c. Major Species in beginning is strong acid d. Major species at equiv. point is the neutral salt bc acid is completely consumed e. Major pH active species at the end is the excess hydroxide from the base f. Endpoint: V1M1 = V2M2 if 1:1 mole ratio g. Multiply (VxMx) by the number of hydrogens the respective compound has to allow for the different mole ratios i. Ex. H2SO4 with 2 Hs and NaOH with 1 OH → 2M1V1 for H2SO4 and 1M1V1 for NaOH iv.

When sketching titration curve, need to find the equivalent point, molarity of acid and thus initial pH, molarity of base and thus final pH (remember to subtract from 14) 1. Initial pH: Adding Acid to Base a. 14-log(Mbase) 2. Initial pH: Adding Base to Acid a. = -log(Macid) 3. Before Equivalence Point pH: Adding Acid to Base: Excess OHa. Determine how many moles of the acid and base b. Determine how many moles of the base remain after the solutions are mixed

c. Determine the total volume d. Determine the concentration of the base e. Determine the pOH f. Convert the pOH to a pH 4. After Equivalence Point pH: Adding Acid to Base: Excess H+ a. Determine how many moles of the acid and base b. Determine how many moles of the acid remain after the solutions are mixed c. Determine the total volume d. Determine the concentration of the acid e. Determine the pH 2. Weak Acid/Base and Strong Base/Acid a. Equivalence Point: Weak Base and Strong Acid i. Moles A = Moles B ii. pH of equiv. Point: Weak base and Strong Acid 1. Find the molarity of the conjugate acid a. Moles of conjugate acid/base is the same as the moles of the base/acid that is the analyte. Find moles by using the given Molarity and Volume b. Find the concentration of the conjugate acid/base by dividing the moles by the TOTAL volume (the number of mL at the equivalence point times 2) 2 2. Ka = x /Mconjugate acid a. Ka found from Kb 3. pH = (-logx) 4. Vtotal = Vanalytye + Vtitrant iii. pH of Equivalence Point: Weak Acid and Strong Base 1. Find the molarity of the base 2. Kb = x2/Mbase 3. pOH = -logx 4. pH = 14 - (-logx) 5. Vtotal = Vanalytye + Vtitrant iv. b. pH of Starting Point: Weak Base and Strong Acid i. Find Kb ii. Kb = x2/Mbase iii. pOH = -logx iv. pH = 14 - (-logx) c. Initial pH is higher than that of strong curve d. Modified S shape curve e. Weak acid doesn’t completely dissociate so fewer hydronium ions at a time than that pf the strong curves f. Beginning of Titration i. pH of weak acid curve = Ka = x2/Mbase. Solve for x and then pH = -logx

g. Not a neutralization reaction but rather acid-base reactions h. Half Equivalence Point when we are 50% of the way to the equivalence point and [A -] = [HA] so [A-]/[HA] = 1 i. Method 1: Convert Ka to pKa 1. pKa = -log(Ka) 2. pH = pKa +log[A-]/[HA] = pKa + log 1 = pKa ii. Method 2 1. Ka = ([H+][A-])/[HA] = [H+] * 1 2. [H+] = Ka 3. pH = pKa iii. BEFORE THE HALF EQUIVALENCE POINT Solve for pH when not at the half equivalence point for by understanding the initial volume and the added volume helps give the proportion [A-]/[HA]. 1. Added Over Original = x% to the equivalence point. If x% is added, then 100-x% = y% remains. 2. x%/y% = [A-]/[HA] 3. pH = pKa +log[A-]/[HA] iv. After the HALF Equivalence Point 1. M1V1 = M2V2 2. M1 is the molarity of the thing you’re adding 3. V1 is how much greater the addition is than the mL of the ½ equivalence point 4. M2 is unknown 5. V2 is the mL of the thing added + the mL at the equivalence point 6. If the thing added is a base, then pH is 14- (-logM2) 7. If the thing added is an acid, then pH is -logM2 v. At the EQUIVALENCE point 1. pH dependent on concentration of conjugate acid/base 2. Moles of conjugate acid/base is the same as the moles of the base/acid that is the analyte. Find moles by using the given Molarity and Volume 3. Find the concentration of the conjugate acid/base by dividing the moles by the TOTAL volume (the number of mL at the equivalence point times 2) 4. Set up the Ka/b = x2/M and solve for x 5. Then find the pH by doing -logx if its a conjugate acid and 14- -logx if its a base 3. Buffer Zone a. Area between initial and buffer point tells difference between weak and strong bases/acid b. Strong acids don’t form buffers so no buffer zone on titration curve c. Both weak acid and conjugate base is present d. Beginning: only weak acid. Conjugate base is soon introduced tho. Conjugate base not necessarily the base being added to solution e. End (the endpoint): weak acid fully turned into conjugate base. Conjugate base is now the major species

f. After end point, major species is only the strong base g. Identify midpoint which is halfway to the endpoint i. Moles HA consumed = Moles A- produced ii. [A-] = [HA] so [A-]/[HA] = 1 iii. Method 1: Convert Ka to pKa 1. pKa = -log(Ka) 2. pH = pKa +log[A-]/[HA] = pKa + log 1 = pKa 3. pH = pKa iv. Method 2 1. Ka = ([H+][A-])/[HA] = [H+] * 1 2. [H+] = Ka v. pKa can then be used to identify the species h. Buffer capacity is not buffer pH range i. Buffer pH range = pKa +/- 1 pH unit ii. Buffer zone = pKa +/- 1 pH unit iii. Buffer capacity is the number of moles (M*V) of added acid/base that buffer can absorb before exceeding the buffer zone 1. Adding acid gets to the lower end of buffer zone 2. Adding base gets to the higher end of buffer zone 3. If buffers equally well against acid/base (added or removed acid/base is equal number) then ideal buffer 4. If can buffer against less base than acid then good acid against acid and not good against base 4. Polyprotic acid-base reactions a. Multiple dissociationa b. Looks like 2 s shaped curve c. As many endpoints as protons it can lose i. H3PO4 has 3 endpoints d. Before adding anything i. Set Ka/b = x2/(M-x) ii. Solve for x iii. pH = 14- -logx or just -logx depending e. At the very end i. Find Ka/b = x2/M 1. Remember Kb = 10-14/Ka 2. Kb if you’re adding the base and Ka if you’re adding the acid ii. Solve for M with M1V1 = M2V2 iii. M1 is the molarity of the thing you’re adding iv. V1 is how much greater the addition is than the mL of the equivalence point v. M2 is unknown vi. V2 is the mL of the thing added + the mL at the equivalence point f. Endpoint 1 = H2PO4- = ½(pKa1+pKa2) i. Only one pH active species which is H2PO4g. Midpoint 1: pH = pKa1

h. Endpoint 2 = HPO4-2 = ½(pKa2+pKa3) i. Only one pH active species which is HPO4-2 i. Midpoint 2: pH = pKa2 j. Endpoint 3 = PO4-3 i. If affected by strong base, then there will be two pH species because the pH isn't high enough to destroy the other pH species ii. Basically looks like a steady line tho, not the standard vertical line iii. Mol base = mol acid k. Midpoint 3 is not pKa3 because Ka2 is very similar to Kw so the equilibria are competing...