Question

Berapakah pH campuran dari 500 ml larutan asam benzoat 0.1 M dan 500 ml Natrium benzoat 0.01 M. (Ka asam benzoat 6.5 x 10^-5)

Answer 1

Diketahui :
Volume Asam benzoat ( C6H5COOH ) = 500 ml = 0,5 L
Molaritas C6H5COOH = 0,1 M = 0,1 mol/L
Volume Natrium benzoat ( C6H5COONa ) = 500 ml = 0,5 L
Molaritas C6H5COONa = 0,01 M = 0,01 mol/L
Ka = 6,5 × 10^ -5

Ditanya : pH campuran ?

Jawab :

n C6H5COOH :
= M × V
= 0,1 mol/L × 0,5 L
= 0,05 mol

n C6H5COONa :
= M × V
= 0,01 mol/L × 0,5 L
= 0,005 mol

[H+] :
= Ka × mol asam / mol garam
= 6,5 × 10^ -5 × 0,05 mol / 0,005 mol
= 6,5 × 10^ -4

pH :
= - log [H+]
= - log 6,5 × 10^ -4
= - ( log 6,5 + log 10^ -4)
= - log 6,5 - log 10^ -4
= 4 - log 6,5