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A step-by-step calculation of ph values at various stages during the titration of a strong acid (0.10 m hcl) with a strong base (0.20 m naoh). The initial point, equivalence point, and points before and after the equivalence point. It also compares the titration curve of a strong acid-strong base system with that of a weak acid-strong base system.
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Titration Curve Review Titration curve - plot of pH vs volume of titrant added For curve - important points to calculate pH and plot are:
-> H 2 O
I (initial) 5 5. C (change) -5 -5 (Limiting Reactant) mol OH - = M X L = 0.2 X 26 mL = 5.2 mmol F (final) 0 0. pOH = -log (0.0.0026) = 2.6; pH = 11.4 (^) M OH-^ = mol/Vtotal = 0.2 mol/76 mL = 0.0026 M
Let's compare Strong Acid/Strong Base Titration with that of Weak Acid/Strong Base (or vice versa) both with 50.0 mL 0.10 M acid with 0.20 M NaOH - Strong acid and strong base
1. Initial point (0 mL of NaOH added) 1. Initial point (0 mL of NaOH added) HCl -> H+^ + Cl-^ HA <-> H+^ + A- 0.1 M -> 0.1 M Create Ice Table, Need Ka, Solve for H+ then pH = -log (0.1) = 1 find pH 2. Equivalence point 2. Equivalence point First let's calculate what volume of NaOH was added <-- Do same to find volume of base but then… using mol H + = mol OH - or M (^) H+ V (^) a = M (^) OH- V (^) b HA^ +^ OH - -> H2O + A - so Vb = (0.10 M)(50 mL)/(0.20 M ) = 25 mL Use ICF table to determine amount of A- produced Since it's a strong acid-strong base titration, at eq pt in moles but then convert to concentration H + + OH - -> H 2 O it's neutral! pH = 7 moles HA and OH- will be 0 since eq pt Finally consider… A- + H2O <-> HA +OH- Create Ice table with A- initial conc, need to calc Kb using Kw/Ka, solve for OH-, then pOH-> pH YIKES!!! 3. Before eq pt, V (^) NaOH = 24 mL 3. Before eq pt, V (^) NaOH = 24 mL Let's use stoichiometry -> moles <-- Do same but with equation… H+^ + OH-^ -> H 2 O HA^ +^ OH - -> H2O + A - I (initial) (^5) 4.8 I C (change) -4.8 -4.8 (Limiting Reactant) C F (final) 0.2 0 F x 0 y pH = -log (0.0.0027) = 2.6 Then use Henderson-Hasselbach with y and x 4. After eq pt, V (^) NaOH = 26 mL 4. After eq pt, V (^) NaOH = 26 mL H + + OH - -> H 2 O <-- Do same but with equation… I (initial) 5 5.2 HA + OH-^ -> H2O + A- C (change) -5 -5 (Limiting Reactant) I F (final) 0 0.2 C F 0 x y pOH = -log (0.0.0026) = 2.6; pH = 11. /\ (0.2/76) Then, can try ICE with A- like you did at end of step 2 but you WILL have an initial amount of OH-. But majority of the OH- will come from the excess, unreacted OH-. So you really can do the same thing you did at the end on the left for strong acid/base