Question

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M acetic acid (Ka = 1.75x10-5) with 0.150 M NaOH(aq). (1 point each) (a) Before addition of any NaOH (b) After addition of 5.0 mL of NaOH (c) After addition of 15.0 mL of NaOH (d) After addition of 25.0 mL of NaOH (e) After addition of 40.0 mL of NaOH (f) After addition of 60 mL of NaOH

Answer 1

A)

Before addition of base pH is due dissociation of weak acid.

Acetic acid is weak acid

pH = $\frac{1}{2}$ ( pKa - logC)

Given, pH = $\frac{1}{2}$ ( - log Ka - log 0.150)

=$\frac{1}{2}$ (-log 1.75×10^-5 - 0.8239)

= $\frac{1}{2}$ ( 4.75 - 0.8239)

= 1.963

B)

Initial mmoles of acetic acid (AcOH) =25.0×0.150 = 3.75

mmoles of NaOH added = (5×0.150) = 0.75

NaOH + AcOH $\rightarrow$ NaAc + H₂O

BCA table is

	AcOH	NaOH	NaAc
Before	3.75	0.75	0
Change	-0.75	-0.75	+0.75
After	3.00	0	0.75

Now, the solution contain excess weak acid and its salt ,hence it is now an acid buffer.

pH = pKa + log $\frac{[NaAc]}{[AcOH]}$

= 4.75 + log (0.75/3.00)

= ( 4.75 - 0.60)

= 4.15

C)

mmoles added = (15×0.150) = 2.25

BCA table is

	AcOH	NaOH	NaAc
Before	3.75	2.25	0
Change	-2.25	-2.25	+2.25
After	1.50	0	2.25

pH = pKa + log (2.25/1.50)

= ( 4.75 + 0.176 )

pH = 4.926

d)

When 25 ml NaOH is added the reaction reached at equivalence point.

mmoles of salt = 25× 0.15 = 3.75

The pH is due to dissociation of the salt .

Now concentration of salt = mmoles of salt/ total volume

= 3.75/(25+25)

= 0.075 M

Now, pH = 7+ $\frac{1}{2}$ (pKa + logC)

= 7 +  $\frac{1}{2}$ ( 4.75 + log 0.075)

= 7 + 1.812 = 8.812

e)

When 40 ml NaOH is added, pH is due to dissociation of strong base NaOH

Now, excess mmoles of base = (40×0.150) - 3.75 = 2.25

Concentration of base = ( mmoles of base/ total volume)

= (2.25/40+25)

= 0.0346 M

pOH = - log (0.0346) = 1.46

0r, pH = 14 -1.46 =12.54

f)

After addition of 60 ml NaOH

Excess moles of NaOH = (60 ×0.150) - 3.75 = 5.25

Then concentration of base = 5.25/(60+25)

= 0.0617 M.

pOH = -log (0.0617) = -1.209

Or, pH =14 - 1.21 = 12.79