Titrations

Introduced to this in AP Chemistry and Grade 11 Chemistry (Rick Hansen Secondary School)

What is it §

In titrations a solution in a buret is gradually added to another solution until the chemical reaction between the two solutions is complete

Equivalence Point - The point when the rxn is complete (calculated)

Indicator - substance that changes color at the endpoint

Endpoint - the approximate time where the color has changed (general)

Indicator - Substance that changes color at endpoint

Example: Strong Acid and Strong Base §

A solution of HCl is titrated with NaOH

Step #1: What is in solution? §

H${}^{+}$ $\cancel{C{l}^{-}}$ $\cancel{N{a}^{+}}OH$^-$

Ignore Cl and Na

Step #2: Write an equation to consume the strong acid/base §

OH${}^{-}$ + H${}^{+}$ $\to$ H${}_2$O

Strong Acid-Strong Base Titrations §

Example: Weak Acid and Strong Base §

A solution of Acetic Acid, HC${}_2$H${}_3$O${}_2$, is titrated with NaOH

Step #1: What is in solution? §

H${}^{+}$ C${}_2$H${}_3$O${}_2$ $\cancel{N{a}^{+}}$ OH${}^{-}$

Step #2: Write an equation to consume the strong base §

OH${}^{-}$ + HC${}_2$H${}_3$O${}_2$ $\to$ C${}_2$H${}_3$O${}_2$ + H${}_2$O

Weak Acid-Strong Base Titrations §

CH${}_3$COOH (aq) + OH${}^{-}$ (aq) $\to$ CH${}_3$COO${}^{-}$ (aq) + H${}_2$O (l)

pH will be slightly above 7 which lets us know weak acid and strong base

Example: Weak Base and Strong Acid §

A solution of ammonia, NH${}_3$, is titrated with HCl

Step #1: What is in solution? §

NH${}_3$ H${}^{+}$ $\cancel{C{l}^{-}}$

Step #2: Write an equation to consume strong acid §

NH${}_3$ + H${}^{+}$ $\to$ NH${}_4^{+}$

Step #3: §

Summary §

• Strong Acid vs. Strong Base
  • 100% ionized pH = 7 No equilibrium
• Weak Acid vs Strong Base
  • Acid is neutralized; Need K${}_b$ for conjugate base equilibrium pH > 7
• Strong Acid vs Weak Base
  • Base is neutralized; Need K${}_a$ for conjugate acid equilibrium pH < 7

Titration Problems - finding the pH of the solution §

1. Write the net ionic equation
2. Stoichiometry problem - react the strong acid/base
   1. Moles (or millimoles = Molarity x mL)
3. Equilibrium, if a weak acid or base is present
   1. ICE table (Molarity)
   2. Henderson-Hasselbach Equation (Molarity)
4. Calculate pH

Titration - strong acid + strong base §

What is the pH at different points during the titration of 50.0 mL of 0.100 M HCl and 0.200 M KOH

Step #1: What is in solution §

H${}^{+}$ $\cancel{C{l}^{-}}$

Step #4: no acid/base reaction §

$\therefore$ pH = -log(0.100) = 1.000

What is the pH at different points during titration of 50.0 mL of 0.100 M HCl and 0.200 M KOH

Step #1: What is in solution §

H${}^{+}$ $\cancel{C{l}^{-}}$ $\cancel{K^{+}}$ OH${}^{-}$

Step #2: Write a net ionic equation to react strong acid/base §

OH${}^{-}$ + H${}^{+}$ $\to$ H${}_2$O

What is the pH at different points during titration of 50.0 mL of 0.100 M HCl and 0.200 M KOH

After addition of 25.0 mL of KOH

Step #1: What is in solution §

H${}^{+}$ $\cancel{C{l}^{-}}$ $\cancel{K^{+}}$ OH${}^{-}$

Step #2: Write a net ionic equation to react strong acid/base §

OH${}^{-}$ + H${}^{+}$ $\to$ H${}_2$O

Step #3: §

Titration - weak acid + strong base §

What is the pH at different points during the titration of 50.0 mL of 0.100 M HCOOH (K${}_a$ = 1.8 * 10${}^{-4}$) and 0.200 M of KOH

A. Before addition of KOH §

Step # 1 §

HCOOH $\cancel{K}$ OH${}^{-}$

Step # 3 §

HCOOH $\leftrightarrow$ H${}^{+}$ + HCOO${}^{-}$

Reactant/Product	Initial	Change	Equilibrium
HCOOH	0.100 M	-x	0.100 - x
H${}^{+}$	0	+x	+ x
HCOO${}^{-}$	0	+x	+x

K${}_a$ = $\frac{[H^{+}][HCO{O}^{-}]}{[HCOOH]}$ = $\frac{x^2}{0.100-x}$ ~ $\frac{x^2}{0.100}$

$\sqrt{1.8\ast 10^{-4}\ast 0.100}$ = x

-log(x) = 2.37

pH = 2.37

B. After the addition of 10.0 mL of KOH added §

Step #1: What is in the solution §

HCOOH $\cancel{K^{+}}$ OH${}^{-}$

Step #2: Net ionic equation §

OH${}^{-}$ + HCOOH $\to$ HCOO${}^{-}$ + H${}_2$O

Reactant/Product	Before rxn	Consuming L.R	After rxn
OH${}^{-}$	(10.0 mL)(0.200M) = 2.00 mmol	-2 mmol	0
HCOOH	(50.0mL)(0.100M) = 5.00 mmol	-2 mmol	3.0 mmol
HCOO${}^{-}$	0	+ 2 mmol	2.00 mmol

pH = pK${}_a$ + log$\frac{[A^{-}]}{[HA]}$ = -log(1.8 * 10${}^{-4}$ )

C. After addition of 12.5 mL of KOH §

Step #1 §

HCOOH $\cancel{K}$ OH

Step #2 §

HCOOH + OH${}^{-}$ $\to$ HCOO${}^{-}$ + H${}_2$O

Reactant/Product	Before rxn	Consuming L.R	After rxn
OH${}^{-}$	(12.5 mL)(0.200M) = 2.50 mmol	-2.5 mmol	0
HCOOH	(50.0mL)(0.100M) = 5.00 mmol	-2.5 mmol	2.5 mmol
HCOO${}^{-}$	0	+ 2.5 mmol	2.50 mmol

pH = -log(1.8 * 10${}^{-4}$) + log$\frac{2.5mmol/62.5}{2.5mmol/62.5}$ = 3.74 + 0 = 3.74

half equivalence point (aka halfway point): pH = pK${}_a$

After the addition of 25.0 mL of KOH §

Step # 1: What is in solution? §

HCOOH $\cancel{K^{+}}$ OH${}^{-}$

Step #2: Write a net ionic equation to react §

OH${}^{-}$ + HCOOH $\to$ HCOO${}^{-}$ + H${}_2$O

Reactant/Product	Before	Consuming L.R	After rxn
OH${}^{-}$	0	-5 mmol	0
HCOOH	0	-5 mmol	0
HCOO${}^{-}$	5 mmol/75 mL = 0.067M	+5 mmol	5 mmol

Step #3: Equilibrium §

HCOO${}^{-}$ + H${}_2$O $\leftrightarrow$ OH${}^{-}$ + HCOOH

0.067 M

K${}_b$ = $\frac{[O{H}^{-}][HCOOH]}{[HCO{O}^{-1}]}$ = $\frac{x^2}{0.067}$ = $\frac{K_w}{1.8\ast 10^{-4}}$ =