Question

a 22.5 ml sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 ml of the base is added. What was the concentration of acetic acid in the original 22.5 ml? what is the ph of the equivalence point? (ka (acetic acid) = 1.75 x 10^-5)

Answer 1

Given that ; (ka (acetic acid) = 1.75 x 10^-5)

CH3COOH + NaOH = CH3COONa+ H2O

Moles of NaOH = Molarity * volume in L

= 0.175* 37.5/1000

= 0.0065625 Moles NaOH

Now calculate the moles CH3COOH as follows:

0.0065625 Moles NaOH * 1 mole CH3COOH/ 1 mole NaOH

= 0.0065625 Moles CH3COOH

Molarity of CH3COOH = Number of moles / volume in L

= 0.0065625 Moles CH3COOH /22.5 ml*1 L /1000 ml

= 0.292 M CH3COOH

The acetic acid is a weak acid while NaOH is a strong base, thus at the equivalent point pH is basic means more than 7.

First calculate the K b of CH3COO- as follows:

Kb = Kw/Ka

= 1.00*10^-14/ 1.75*10^-5

= 5.7*10^-10

Now calculat the concnetrtaion of [OH-] as follows:

[OH-] = √Kb *C

= √5.7*10^-10 *0.292

= √1.6644*10^10

= 1.290*10^-5 M

pOH = - log [OH-]

= - log 1.290*10^-5

= 4.89

We know that pH + pOH = 14

Then pH = 14-pOH

= 14- 4.89

= 9.11