Titration Curves What’s in the beaker. Always ask… What’s in the beaker? When I start the...

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 0.100 M HCl diluted with 50.0 mL of water


Cl-

H2O

Cl-

H2O

Cl-

H2O

H+

H+

H+

I know what’s in the beaker…

Now I ask: what reactions are possible?

Cl-

H2O

Cl-

H2O

Cl-

H2O

H+

H+

H+

“What reactions are possible”?

H+ + H2O H+ + H2O

Cl- + H2O HCl + OH-

H2O + H2O H3O+ + OH-

Which reactions matter?

NONE!

Cl-

H2O

Cl-

H2O

Cl-

H2O

H+

H+

H+


H+ + H2O H+ + H2O

Cl- + H2O HCl + OH-


10.00 mL of 0.100 M HCl diluted with 50.0 mL of water

What’s the pH?

0.100 M * 10.0 mL = [HCl]* 60.0 mL

[HCl] = 1.667x10-2 M = [H+] = [Cl-] (it’s a strong acid)

pH = - log (1.667x10-2 M) = 1.778

10.00 mL of 0.100 M HCl diluted with 50.0 mL of water. Then add 2.00 mL of 0.100 M

NaOH.What’s in the beaker?

Cl-

H2O

Cl-

H2O

Cl-

H2O

H+

Na+

H+



Cl-

H2O

Cl-

H2O

Cl-

H2O

Na+

H+

H+


H+ + H2O H+ + H2O

Cl- + H2O HCl + OH-

Na+ + H2O NaOH + H+



NONE!

Cl-

H2O

Cl-

H2O

Cl-

H2O

H+

H+

H+


H+ + H2O H+ + H2O

Cl- + H2O HCl + OH-

Na+ + H2O NaOH + H+



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.

0.100 M * 0.0100 L = 0.00100 moles HCl initial

0.100 M NaOH * 0.00200 L = 0.0002 mol NaOH

Always start with a balanced equation. Which one?

H+ + OH- H2O

I 0.001 mol 0.0002 mol -

C -0.0002 mol -0.0002 mol -

End 0.0008 mol 0 mol

So….

0.0008 mol H+ = 1.29x10-2 M

0.062 L

pH = - log(1.29x10-2 M) = 1.889

Notice, not a lot of change.



things!



0.100 M NaOH * 0.00500 L = 0.0005 mol NaOH


H+ + OH- H2O

I 0.001 mol 0.0005 mol -

C -0.0005 mol -0.0005 mol -

End 0.0005 mol 0 mol

So….

0.0005 mol H+ = 7.69x10-3 M

0.065 L

pH = - log(7.69x10-3 M) = 2.114

Notice, not a lot of change.



things!



0.100 M NaOH * 0.0100 L = 0.00100 mol NaOH


H+ + OH- H2O

I 0.001 mol 0.001 mol -

C -0.001 mol -0.001 mol -

End 0.00 mol 0 mol

So….

Nothing left….so is [H+] = 0?

Suddenly, one of my reactions is relevant:


The pH is 7 at equivalence because of the Kw

reaction.


NaOH.What’s in the beaker?

Cl-

H2O

Cl-

H2O

Cl-

H2O

Na+

Na+

Na+

Na+

OH-

OH-

OH-

Na+

Na+




OH- + H2O OH- + H2O

Cl- + H2O HCl + OH-

Na+ + H2O NaOH + H+


Cl-

H2O

Cl-

H2O

Cl-

H2O

Na+

Na+

Na+



things!



0.100 M NaOH * 0.01200 L = 0.0012 mol NaOH


H+ + OH- H2O

I 0.001 mol 0.0012 mol -

C -0.001 mol -0.001 mol -

End 0.00 mol 0.0002 mol

So….

0.0002 mol OH- = 2.778x10-3 M OH-

0.072 L

pOH = - log(2.778x10-3 M) = 2.556

pH = 14 – 2.556 = 11.444



things!



0.100 M NaOH * 0.01500 L = 0.0015 mol NaOH


H+ + OH- H2O

I 0.001 mol 0.0015 mol -

C -0.001 mol -0.001 mol -

End 0.00 mol 0.0005 mol

So….

0.0005 mol OH- = 6.667x10-3 M OH-

0.075 L

pOH = - log(6.667x10-3 M) = 2.176

pH = 14 – 2.176 = 11.824

10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water


H2O

HOAc

H2O H2O

HOAc

HOAc




HOAc + H2O OAc- + H3O+


H2O

HOAc

H2O H2O

HOAc

HOAc



H2O

HOAc

H2O H2O

HOAc

HOAc

10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water

What’s the pH?

0.100 M * 10.0 mL = [HOAc]* 60.0 mL

[HOAc] = 1.667x10-2 M(initial concentration)

ICE-ICE-BABY-ICE-ICE



I 1.667x10-2 M - 0 0

C -x - +x +x

E 1.667x10-2 -x - x x

So….

Ka = 1.80x10-5 = (x)(x)

(1.667x10-2 M –X)

Assume x<<1.667x10-2

Ka = 1.80x10-5 = (x)(x)

(1.667x10-2 M)

3.000x10-7 = x2

X= 5.48x10-4 M (works…barely)

pH = - log (5.48x10-4) = 3.26


I 1.667x10-2 M - 0 0

C - 5.48x10-4 M - 5.48x10-4 M 5.48x10-4 M

E 1.612x10-2 -x - 5.48x10-4 M 5.48x10-4 M

10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water. Add 2.00

mL NaOH.What’s in the beaker?

OAc-

H2O

HOAc

H2OOAc-

H2O

Na+

HOAc

Na+HOAc

HOAc

Why?

NaOH + HOAc NaOAc + H2O






OAc- + H2O HOAc + OH- (but this is the same as the first one)

OAc-

H2O

HOAc

H2OOAc-

H2O

H+

HOAc

H+HOAc

HOAc



OAc-

H2O

HOAc

H2OOAc-

H2O

H+

HOAc

H+HOAc

HOAc

10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water. Add 2.00 mL

of NaOH.What’s the pH?

0.100 M * 0.010 L = 0.001 mol HOAc

0.100 M * 0.002 L = 0.0002 mol NaOH

DOUBLE ICE- DOUBLE ICE-BABY-DOUBLE ICE- DOUBLE ICE


HOAc + NaOH NaOAc + H2O

I 0.0010 mol

0.0002 mol 0 -

C -0.0002 mol

-0.0002 mol

+0.0002 -

E 0.0008 mol

0 mol 0.0002

On to the 2nd ICE chart


I -

C -x - +x +x

E -



I 0.0008 mol

0.062 L- 0.0002

0.062

C -x - +x +x

E -

THIS IS ACTUALLY A BUFFER!!!

So….H-H equation

Ka = 1.80x10-5 =

pKa = - log (1.80x10-5) = 4.74

pH = 4.74 + log [OAc-] [HOAc]

pH = 4.74 + log (0.0002) = 4.14 (0.0008)



OAc-

H2O

HOAc

H2OOAc-

H2O

Na+

HOAc

Na+HOAc

HOAc

Why?







OAc- + H2O HOAc + OH- (but this is the same as the first one)

OAc-

H2O

HOAc

H2OOAc-

H2O

Na+

HOAc

Na+HOAc

HOAc



OAc-

H2O

HOAc

H2OOAc-

H2O

Na+

HOAc

Na+HOAc

HOAc

10.00 mL of 0.100 M HOAc diluted with 50.0 mL of water. Add 5.00 mL

of NaOH.What’s the pH?

0.100 M * 0.010 L = 0.001 mol HOAc

0.100 M * 0.005 L = 0.0005 mol NaOH




I 0.0010 mol

0.0005 mol 0 -

C -0.0005 mol

-0.0005 mol

+0.0005 -

E 0.0005 mol

0 mol 0.0005



I 0.0005 - 0.0005

C -x - +x +x

E -

So….H-H equation

Ka = 1.80x10-5 =

pKa = - log (1.80x10-5) = 4.74

pH = 4.74 + log [OAc-] [HOAc]

pH = 4.74 + log (0.0005) = 4.74 (0.0005)



OAc-

H2O

OAc-

H2OOAc-

H2O

Na+

OAc-

Na+OAc-

Na+

Why?






OAc- + H2O HOAc + OH-

No more HOAc, no more Ka

OAc-

H2O

Na+

H2OOAc-

H2O

Na+

OAc-

Na+OAc

Na+


OAc- + H2O HOAc + OH-


mL of NaOH.What’s the pH?

0.100 M * 0.010 L = 0.001 mol HOAc

0.100 M * 0.010 L = 0.001 mol NaOH




I 0.0010 mol

0.0005 mol 0 -

C -0.0010 mol

-0.0010 mol

+0.001 -

E 0 0 mol 0.001


OAc + H2O HOAc + OH-

I 0.001 mol

0.070 L

- 0. 0

C -x - +x +x

E 6.9x10-2 -x - x x

So….

Kb = Kw = 1.0x10-14 = 5.56x10-10 Ka 1.80x10-5

5.56x10-10 = (x)(x) 0.069-XAssume x<<0.069x10-2

5.56x10-10 = (x)(x) 0.069

3.836x10-11 = x2

X= 6.19x10-6 M

pOH = - log 6.19x10-6 = 5.2pH = 14 – 5.2 = 8.8

OAc + H2O HOAc + OH-

I 0.001 mol

0.070 L

- 0. 0

C -6.19x10-6 M - +6.19x10-6 M +6.19x10-6 M

E 6.9x10-2 - 6.19x10-6 M 6.19x10-6 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!

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