• Calculate pH of a solution containing 25.00 mL of 0.12 M formic acid (Ka = 1.7x10−4) to which the following amounts of 0.10 M NaOH have been added:

o 0.00 mL – STARTING PT: regular ICE table

Ice Table this type of arrow b/c equilibrium

(MOLARITY) HCOOH + H2O ⇌ H3O+ + HCOO− I 0.12 M − 0 0 C −x − +x +x E 0.12−x − x x 𝐾! = 1.7×10!! = !"#!! [!!!!]

[!"##!]

1.7×10!! =𝑥!

0.12 − 𝑥

B/c [!"##!]!"!#!!

= !.!"!.!×!"!!

> 400, lose the x in any expressions

1.7×10!! =x!

0.12

⇒ 𝑥 = 0.0045 H!O! = x ∴ pH = − log x = 2.35

o 15.0 mL – BEFORE EQ PT ! Method 1: titration table (TT) & HH equation

• This method only works in buffer region (b/w starting pt. and eq. pt.)

Total volume: 40 mL HCOOH: n!"##! = 0.0030  mol NaOH: n!"#$ = 0.0015  mol

HCOOH = !.!!"!!"!#$  !"#.

= 0.075  M HCOOH = !.!!"#!"!#$  !"#.

= 0.0375  M

What’s in the beaker after adding the NaOH – values for “I” row if using method 2

Titration Table this type of arrow b/c goes to completion

(MOLES) HCOOH + OH− " HCOO− + H2O ba 0.0030 0 0 − a 0 0.0015 0 − aa 0.0015 0 0.0015 − pH = p𝐾! + log

!!"#$.!"#$!!"#$  !"#$

= − log 1.7×10!! + log0.00150.0015

= 3.77 + log 1 = 3.77

25 mL 0.12 M 15 mL 0.10 M

HCOOH = 0.0015

H3O+ = 0 H2O

HCOO− = 0.0015 Shelby

! Method 2: TT & ICE table (Ka) • Use TT above, then use “aa” values & total vol. for “I” row

Ice Table

this type of arrow b/c equilibrium (MOLARITY) HCOOH + H2O ⇌ H3O+ + HCOO− I !.!!"#  !"#

!.!"!  ! = 0.0375  M − 0 !.!!"#

!.!"!= 0.0375  M

C −x − +x +x E 0.0375−x − x 0.0375+x 𝐾! = 1.7×10!! = !"#!! [!!!!]

[!"##!]= !.!"#$!! !

(!.!"#$!!)⇒ 𝑥 = 0.000168

H!O! = x ∴ pH = − log x = 3.77

o 30.0 mL – @ EQ PT: TT & ICE table (Kb )

Titration Table HCOOH + OH− " HCOO− + H2O

ba 0.0030 0 0 − a 0 0.0030 0 − aa 0 0 0.0030 −

Ice Table HCOO− + H2O ⇌ HCOOH + OH−

I !.!!"!  !"#!.!""  !

= 0.054  M − 0 0 C −x − +x +x E 0.054− x − x x 𝐾! =

10!!"  (𝐾!)1.7×10!!  (HCOOH  𝐾!)

= 5.88×10!!! =OH! [HCOOH][HCOO!]

5.88×10!!! =x!

0.054 − x

B/c [!"#!!]!"!#

!!= !.!"!

!.!!×!"!!!> 400, lose the x in any expressions

5.88×10!!! =x!

0.054

⇒ x = 1.79×10!! OH! = x ∴ pOH = − log x = 5.75 ∴ pH = 14 − 5.75 = 8.25

o 31.0 mL – AFTER EQ PT: TT [OH−]

Titration Table

HCOOH + OH− " HCOO− + H2O ba 0.0030 0 0 − a 0 0.0031 0 − aa 0 0.0001 0.0030 −

Shelby

! Find leftover nNaOH, value in “aa” row, then calculate [OH−] • Leftover nNaOH = 0.0001 • [OH−] = 0.001 mol/0.056 L (total volume) = 0.001786 M

! Calculate pOH then pH • pOH = − log(0.001786  ) = 2.75 ∴ pH = 14 − 2.75 = 11.25

• Things to Remember for Titration Problems

o There are 4 diff. “areas” – starting pt, buffer region, eq pt, and after eq pt o @ eq pt: [HA] = 0, A− =   !!"#$%#&'

!!".!"., and nHA=nB; @ ½ eq pt: pH = pKa

o Strong acid/strong base titration: eq. pt. pH=7 o Weak acid/strong base titration: eq. pt. pH>7 o Weak base/strong acid titration: eq. pt. pH<7 o Each proton titrated separately in titrations involving polyprotic acids

Shelby