Titration Curves What’s in the beaker. Always ask… What’s in the beaker? When I start the titration, there is only water and whatever it is I’m titrating.

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Presentation transcript:

Titration Curves What’s in the beaker

Always ask… What’s in the beaker? When I start the titration, there is only water and whatever it is I’m titrating (and maybe indicator, but we’ll ignore that). Let’s titrate HCl with NaOH

10.00 mL of M HCl diluted with 50.0 mL of water What’s in the beaker? Cl - H2OH2O H2OH2O H2OH2O H+H+ H+H+ H+H+

I know what’s in the beaker… Now I ask: what reactions are possible? Cl - H2OH2O H2OH2O H2OH2O H+H+ H+H+ H+H+ “What reactions are possible”? H + + H 2 O  H + + H 2 O Cl - + H 2 O  HCl + OH - H 2 O + H 2 O  H 3 O + + OH -

Which reactions matter? NONE! Cl - H2OH2O H2OH2O H2OH2O H+H+ H+H+ H+H+ “What reactions are possible”? H + + H 2 O  H + + H 2 O Cl - + H 2 O  HCl + OH - H 2 O + H 2 O  H 3 O + + OH -

10.00 mL of M HCl diluted with 50.0 mL of water What’s the pH? M * 10.0 mL = [HCl]* 60.0 mL [HCl] = x10 -2 M = [H + ] = [Cl - ] (it’s a strong acid) pH = - log ( x10 -2 M) = 1.778

10.00 mL of M HCl diluted with 50.0 mL of water. Then add 2.00 mL of M NaOH. What’s in the beaker? Cl - H2OH2O H2OH2O H2OH2O H+H+ Na + H+H+

I know what’s in the beaker… Now I ask: what reactions are possible? Cl - H2OH2O H2OH2O H2OH2O Na + H+H+ H+H+ “What reactions are possible”? H + + H 2 O  H + + H 2 O Cl - + H 2 O  HCl + OH - Na + + H 2 O  NaOH + H + H 2 O + H 2 O  H 3 O + + OH -

Which reactions matter? NONE! Cl - H2OH2O H2OH2O H2OH2O H+H+ H+H+ H+H+ “What reactions are possible”? H + + H 2 O  H + + H 2 O Cl - + H 2 O  HCl + OH - Na+ + H2O  NaOH + H+ H 2 O + H 2 O  H 3 O + + OH -

10.00 mL of M HCl diluted with 50.0 mL of water. Then add 2.00 mL of M NaOH. What’s the pH? I like an ICE chart here, just to keep track of things! Moles is better for my ICE chart because of dilution issues M * L = moles HCl initial M NaOH * L = mol NaOH

Always start with a balanced equation. Which one? H++OH -  H2OH2O I0.001 mol mol- C mol - End mol0 mol

So… mol H + = 1.29x10 -2 M L pH = - log(1.29x10 -2 M) = Notice, not a lot of change.

10.00 mL of M HCl diluted with 50.0 mL of water. Then add 5.00 mL of M NaOH. What’s the pH? I like an ICE chart here, just to keep track of things! Moles is better for my ICE chart because of dilution issues M * L = moles HCl initial M NaOH * L = mol NaOH

Always start with a balanced equation. Which one? H++OH -  H2OH2O I0.001 mol mol- C mol - End mol0 mol

So… mol H + = 7.69x10 -3 M L pH = - log(7.69x10 -3 M) = Notice, not a lot of change.

10.00 mL of M HCl diluted with 50.0 mL of water. Then add mL of M NaOH. What’s the pH? I like an ICE chart here, just to keep track of things! Moles is better for my ICE chart because of dilution issues M * L = moles HCl initial M NaOH * L = mol NaOH

Always start with a balanced equation. Which one? H++OH -  H2OH2O I0.001 mol - C mol - End0.00 mol0 mol

So…. Nothing left….so is [H + ] = 0? Suddenly, one of my reactions is relevant: H 2 O + H 2 O  H 3 O+ + OH- The pH is 7 at equivalence because of the K w reaction.

10.00 mL of M HCl diluted with 50.0 mL of water. Then add mL of M NaOH. What’s in the beaker? Cl - H2OH2O H2OH2O H2OH2O Na + OH - Na +

I know what’s in the beaker… Now I ask: what reactions are possible? “What reactions are possible”? OH - + H 2 O  OH - + H 2 O Cl - + H 2 O  HCl + OH - Na + + H 2 O  NaOH + H + H 2 O + H 2 O  H 3 O + + OH - Cl - H2OH2O H2OH2O H2OH2O Na +

10.00 mL of M HCl diluted with 50.0 mL of water. Then add mL of M NaOH. What’s the pH? I like an ICE chart here, just to keep track of things! Moles is better for my ICE chart because of dilution issues M * L = moles HCl initial M NaOH * L = mol NaOH

Always start with a balanced equation. Which one? H++OH -  H2OH2O I0.001 mol mol- C mol - End0.00 mol mol

So… mol OH - = 2.778x10 -3 M OH L pOH = - log(2.778x10 -3 M) = pH = 14 – =

10.00 mL of M HCl diluted with 50.0 mL of water. Then add mL of M NaOH. What’s the pH? I like an ICE chart here, just to keep track of things! Moles is better for my ICE chart because of dilution issues M * L = moles HCl initial M NaOH * L = mol NaOH

Always start with a balanced equation. Which one? H++OH -  H2OH2O I0.001 mol mol- C mol - End0.00 mol mol

So… mol OH - = 6.667x10 -3 M OH L pOH = - log(6.667x10 -3 M) = pH = 14 – =

10.00 mL of M HOAc diluted with 50.0 mL of water What’s in the beaker? H2OH2O HOAc H2OH2O H2OH2O

I know what’s in the beaker… Now I ask: what reactions are possible? “What reactions are possible”? HOAc + H 2 O  OAc - + H 3 O + H 2 O + H 2 O  H 3 O + + OH - H2OH2O HOAc H2OH2O H2OH2O

Which reactions matter? HOAc + H 2 O  OAc- + H 3 O+ H2OH2O HOAc H2OH2O H2OH2O

10.00 mL of M HOAc diluted with 50.0 mL of water What’s the pH? M * 10.0 mL = [HOAc]* 60.0 mL [HOAc] = x10 -2 M (initial concentration) ICE-ICE-BABY-ICE-ICE

Always start with a balanced equation. Which one? HOAc+H2OH2O  OAc - +H3O+H3O+ I x10 -2 M -00 C-x-+x E x x -xx

So…. K a = 1.80x10 -5 = (x)(x) ( x10 -2 M –X) Assume x<<1.667x10 -2 K a = 1.80x10 -5 = (x)(x) ( x10 -2 M) 3.000x10 -7 = x 2 X= 5.48x10 -4 M (works…barely)

pH = - log (5.48x10 -4 ) = 3.26 HOAc+H2OH2O  OAc - +H3O+H3O+ I x10 -2 M -00 C x10 -4 M- 5.48x10 -4 M E 1.612x x-5.48x10 -4 M

10.00 mL of M HOAc diluted with 50.0 mL of water. Add 2.00 mL NaOH. What’s in the beaker? OAc - H2OH2O HOAc H2OH2O OAc - H2OH2O Na + HOAc Na + HOAc Why? NaOH + HOAc  NaOAc + H 2 O

I know what’s in the beaker… Now I ask: what reactions are possible? “What reactions are possible”? HOAc + H 2 O  OAc - + H 3 O + H 2 O + H 2 O  H 3 O + + OH - OAc - + H 2 O  HOAc + OH - (but this is the same as the first one) OAc - H2OH2O HOAc H2OH2O OAc - H2OH2O H+H+ HOAc H+H+

Which reactions matter? HOAc + H 2 O  OAc- + H 3 O+ OAc - H2OH2O HOAc H2OH2O OAc - H2OH2O H+H+ HOAc H+H+

10.00 mL of M HOAc diluted with 50.0 mL of water. Add 2.00 mL of NaOH. What’s the pH? M * L = mol HOAc M * L = mol NaOH DOUBLE ICE- DOUBLE ICE-BABY- DOUBLE ICE- DOUBLE ICE

Always start with a balanced equation. Which one? HOAc+NaOH  NaOAc+H2OH2O I mol mol0- C mol E mol 0 mol0.0002

On to the 2 nd ICE chart HOAc+H2OH2O  OAc - +H3O+H3O+ I- C-x-+x E-

On to the 2 nd ICE chart HOAc+H2OH2O  OAc - +H3O+H3O+ I mol L C-x-+x E-

THIS IS ACTUALLY A BUFFER!!!

So….H-H equation K a = 1.80x10 -5 = pK a = - log (1.80x10 -5 ) = 4.74 pH = log [OAc - ] [HOAc] pH = log (0.0002) = 4.14 (0.0008)

10.00 mL of M HOAc diluted with 50.0 mL of water. Add 5.00 mL NaOH. What’s in the beaker? OAc - H2OH2O HOAc H2OH2O OAc - H2OH2O Na + HOAc Na + HOAc Why? NaOH + HOAc  NaOAc + H 2 O

I know what’s in the beaker… Now I ask: what reactions are possible? “What reactions are possible”? HOAc + H 2 O  OAc - + H 3 O + H 2 O + H 2 O  H 3 O + + OH - OAc - + H 2 O  HOAc + OH - (but this is the same as the first one) OAc - H2OH2O HOAc H2OH2O OAc - H2OH2O Na + HOAc Na + HOAc

Which reactions matter? HOAc + H 2 O  OAc- + H 3 O+ OAc - H2OH2O HOAc H2OH2O OAc - H2OH2O Na + HOAc Na + HOAc

10.00 mL of M HOAc diluted with 50.0 mL of water. Add 5.00 mL of NaOH. What’s the pH? M * L = mol HOAc M * L = mol NaOH DOUBLE ICE- DOUBLE ICE-BABY- DOUBLE ICE- DOUBLE ICE

Always start with a balanced equation. Which one? HOAc+NaOH  NaOAc+H2OH2O I mol mol0- C mol E mol 0 mol0.0005

On to the 2 nd ICE chart HOAc+H2OH2O  OAc - +H3O+H3O+ I C-x-+x E-

So….H-H equation K a = 1.80x10 -5 = pK a = - log (1.80x10 -5 ) = 4.74 pH = log [OAc - ] [HOAc] pH = log (0.0005) = 4.74 (0.0005)

10.00 mL of M HOAc diluted with 50.0 mL of water. Add mL NaOH. What’s in the beaker? OAc - H2OH2O H2OH2O H2OH2O Na + OAc- Na + OAc - Na+ Why? NaOH + HOAc  NaOAc + H 2 O

I know what’s in the beaker… Now I ask: what reactions are possible? “What reactions are possible”? H 2 O + H 2 O  H 3 O + + OH - OAc - + H 2 O  HOAc + OH - No more HOAc, no more K a OAc - H2OH2O Na+ H2OH2O OAc - H2OH2O Na + OAc- Na + OAc Na+

Which reactions matter? OAc - + H 2 O  HOAc + OH-

10.00 mL of M HOAc diluted with 50.0 mL of water. Add mL of NaOH. What’s the pH? M * L = mol HOAc M * L = mol NaOH DOUBLE ICE- DOUBLE ICE-BABY- DOUBLE ICE- DOUBLE ICE

Always start with a balanced equation. Which one? HOAc+NaOH  NaOAc+H2OH2O I mol mol0- C mol E00 mol0.001

On to the 2 nd ICE chart OAc+H2OH2O  HOAc+OH - I0.001 mol L -0.0 C-x-+x E 6.9x x -xx

So…. K b = K w = 1.0x = 5.56x K a 1.80x x = (x)(x) X Assume x<<0.069x x = (x)(x) x = x 2 X= 6.19x10 -6 M

pOH = - log 6.19x10 -6 = 5.2 pH = 14 – 5.2 = 8.8 OAc+H2OH2O  HOAc+OH - I0.001 mol L -0.0 C -6.19x10 -6 M-+6.19x10 -6 M E 6.9x x10 -6 M

So, for titrations in general… In general, titrations do not have endpoints at pH=7. That only occurs for a strong acid and a strong base. If either the acid or base is weak, then there is a conjugate acid and/or conjugate base floating around at equivalence!