Question

A 52.0-ml volume of 0.35M CH3COOH (Ka= 1.8*10^-5) is titrated with 0.40M NaOH . Calculate the pH after the addition of 21.0ml of NaOH.

Answer 1

To calculate the pH after the addition of NaOH to CH3COOH, consider the reaction between the two compounds and use the equation pH = pKa + log([A-]/[HA]).

Step-by-step explanation:

The pH after the addition of 21.0 mL of NaOH can be calculated by considering the reaction between CH3COOH (acetic acid) and NaOH (sodium hydroxide). The balanced chemical equation for the reaction is:

CH3COOH + NaOH → CH3COONa + H2O

Since acetic acid is a weak acid, it does not fully dissociate in water, and its Ka value is given as 1.8 x 10^-5. However, NaOH is a strong base and dissociates completely in water. By using the equation:

pH = pKa + log([A-]/[HA])

we can calculate the pH after the addition of NaOH.

Answer 2

Ka= 1,8•10^-5
pKa = - log(Ka) = 4,74


Equivalence volume :

Ca x Va = Cb x Vb

Vb = Ca x Va / Cb
Vb = 0,35 x 52,0 / 0,40
Vb = 45,5 mL


21,0 mL < 45,5 mL

. . . then :


pH = pKa / 2 - log( (Ca x Va - Cb x V) / (Va + V)) / 2
pH = pKa / 2 - log( (0,35 x 52 - 0,40 x 21) / (52 + 21)) / 2
pH = 3,24