The reaction of acetic acid with sodium hydroxide is:
[tex]CH_3COOH + NaOH CH_3COONa + H_2O[/tex]
The ratio of A-/HA is calculated as follows:
According to Henderson Hasslebach equation:
[tex][A-]/[HA] = 10^p^H^-^p^K^_a[/tex]
[tex]= 10^4^.^7^6^-^4^.^5^ = ^2^.^1^8[/tex]
The total concentration of HA and A- = 2.0 L * 0.25 M = 0.5 mol.
[tex][ A- ]+ [ HA ]= 0.5[/tex]
[tex][ A^- ] = 0.5 – [ HA][/tex]
[ HA] = 0.156 mol and [ A- ]= 0.343 mol
Total of 0.5 moles of acetic acid is required:
[tex]0.5 mol HA * 60.0 g[/tex]
HA = 30.0 g acetic acid
Conversion of 0.343 moles of the acetic acid to acetate can be performed by adding NaOH
[tex]0.343 mol HA * (1 mol NaOH/1 mol HA) * (1000 mL/1 mol NaOH)[/tex]
= 343 mL
Thus, 343 mL of 1M NaOH is required
